Answer:
The answer is 15,000 NExplanation:
To find the force given the mass , velocity and time can be found by using the formula
[tex]f = \frac{m \times v}{t} \\ [/tex]
where
m is the mass
v is the velocity
t is the time
From the question
m = 1500 kg
v = 30 m/s
t = 3 s
We have
[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]
We have the final answer as
15,000 NHope this helps you
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.
Answer:
The number of turns in the solenoid is 22366.
Explanation:
The number of turns in the solenoid can be found using the following equation:
[tex] B = \mu_{0} I\frac{N}{L} [/tex]
Where:
B: is the magnetic field = 8.90 T
L: is the solenoid's length = 0.300 m
N: is the number of turns =?
I: is the current = 95 A
μ₀: is the magnetic constant = 4π×10⁻⁷ H/m
By solving equation (1) for N we have:
[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]
Therefore, the number of turns in the solenoid is 22366.
I hope it helps you!
A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?
Answer:
I believe the electromagnetic field should be reversed.
Explanation:
When a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.
What is an Electromagnetic wave?An electromagnetic wave may be defined as a type of wave that is significantly created as a result of vibrations between an electric field and a magnetic field. These waves are composed of oscillating magnetic and electric fields.
According to the context of this question, when an individual is constructing an electromagnetic wave and then reverses the direction of the current, it will eventually affect the direction of the magnetic field in the same direction with respect to the current. So, if the direction of the current is reversed, the direction of the magnetic field would also be reversed.
Therefore, when a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.
To learn more about Magnetic fields, refer to the link:
https://brainly.com/question/14411049
#SPJ6
Your question seems incomplete. The most probable complete question is as follows:
A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?
The direction of the magnetic field will be reversed. The magnetic field will expand.The magnetic field would be canceled out and disappear.The magnetic field will cause the voltage of the battery to be reduced.If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
This question is incomplete, the complete question is;
A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
Answer: the time constant of the damped oscillation is 47.44s
Explanation:
Given that;
t = 5.0s
Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao
EXPRESSION for amplitude is A(t) = Ao e^-t / T
t is time while T is time constant
so
0.9Ao = Ao e^-t / T
0.9 = e^ -t/T
So we take the natural log of both the sides
ln (0.9) = -t/T
-0.1054 = -t/T
0.1054 = t/T
WE now substitute our value of t
0.1054 = t/T
0.1054T = 5.0
T = 5 / 0.1054
T = 47.44s
therefore the time constant of the damped oscillation is 47.44s
Which of the organisms in the food web above is the top level carnivore
Answer:
apex consumers
Explanation:
they are top
i need help, for physics
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object?
Answer:
The net force on an object is the total force applied on the object after adding up all the forces
In the given diagram,
we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction
Now, we are left with a force of 2N and 10N,
the net force will be the difference of these forces:
Net force = 10N - 2N
Net force = 8N downwards
Another way to do it:
The two 4N forces will be cancelled out,
and we are left with a 2N and a 10N force
(notice how we cancelled equal and opposite forces for the 4N)
We can divide the 10N force into (2N + 8N)
Since the 2N forces are equal and opposite, they will be cancelled out
and we will be left with a net force of 8N downwards
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1
Answer:
The correct option is B
Explanation:
From the question we are told that
The mass of the pile is M
The height is H = 4 y
The vertical distance achieve during the first lift is [tex]h_1 = 3 y[/tex]
The time taken is [tex]t_1 = 4T [/tex]
The vertical distance achieve during the second lift is [tex]h_2 = y[/tex]
The time taken is [tex] t_2 = T [/tex]
Generally the velocity of the crane during the first lift is
[tex]v _1 = \frac{h_1}{t_1 }[/tex]
=> [tex]v _1 = \frac{3 y}{4T }[/tex]
Generally the velocity of the crane during the second lift is
[tex]v _2 = \frac{h_2}{t_2 }[/tex]
=> [tex]v _2 = \frac{ y}{T}[/tex]
Generally the power generated by the crane during the first lift is
[tex]P_1 = F_1 * v_1[/tex]
Here [tex]F_1[/tex] is the weight of the brick which is mathematically represented as
[tex]F_1 = M * g [/tex] , g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{3y}{4T}[/tex]
Generally the power generated by the crane during the first lift is
[tex]P_1 = F_2 * v_2[/tex]
Here [tex]F_2[/tex] is the weight of the brick which is mathematically represented as
[tex]F_2 = M * g [/tex] , g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{y}{T}[/tex]
The ratio of the first power generated to the second power is
[tex]\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T} }{ Mg * \frac{y}{T} }[/tex]
=> [tex]\frac{P_1}{P_2} = \frac{3}{4}[/tex]
=> [tex]P_2 = \frac{4}{3} P_1[/tex]
Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
how are s waves and p waves simuliar?
A.they shake the ground
B.they travel through liquids
C. they arrive at the same time
D.they shake the ground from side to side
Answer:
A
Explanation:
hope this helps
Which best explains a difference between Einstein’s general theory of relativity and his special theory of relativity?
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
His general theory includes uniform and accelerated motion, but his special theory applies only to accelerated motion.
His general theory applies only to accelerated motion, but his special theory includes uniform and accelerated motion.
His general theory applies only to uniform motion, but his special theory includes uniform and accelerated motion.
Answer:
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
Explanation:
According to Einstein's 1915 general theory of relativity, the force of gravity arises from the curvature of space and time.
According to theory of special relativity:
1. The laws of physics are the same for all non-accelerating observers
2. The speed of light in a vacuum was independent of the motion of all observers.
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
Answer:
for those who dont like to read
the answer is A.
hope i helped
Explanation:
Why wouldn't carbon dating work to determine the age of the earth?
A) Carbon dating works best on other planets
B) The half life of carbon is too short
C) The age of the earth cannot be determined
D) The half life of carbon is too long.
Answer:
The half-life of carbon is too short.
Explanation:
The answer is B.
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
You are hired to lift 30 kg crates a vertically 0.90 m from the ground onto a truck. How many crates would you have to load onto the truck in 1 minute for your average power output in lifting the crates to be 100 W
Answer:
22 crates
Explanation:
Power = Force ×Distance/time taken
Power = m×a×d/t
Power = 30×9.81×0.9/60 (1min was converted to second)
Power = 264.87/60
Power = 4.4145Watts
If my average power output us 100aw, then;
Number of crate to load = average power/4.4145
Number of crates to load = 100/4.4146
Number of crates to load = 22.6
Hence I will have to load about 22 crates onto the truck in 1 minute for my average power output in lifting the crates to be 100 W.
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will
Complete Question:
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will
A. be behind the package.
B. be over the package.
C. be in front of the package.
D. depend on the speed of the plane when the package was released.
Answer:
B.
Explanation:
As no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed, so when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.
The package hits the ground in such a way that the plane will be exactly over the package.
The given problem is based on the effect of air resistance on the motion of object. The resistive force exerted on any object on account of dust, and other air particles is known as air resistance or simply as air drag.
In the given problem, as no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed.
So when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.
Thus, we can conclude that the package hits the ground in such a way that the plane will be exactly over the package.
Learn more about the air drag here:
https://brainly.com/question/14755232
Please provide an explanation.
Thank you!!
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kNb. Left support = 36 kN Right support = 29 kNc. Left support force will decrease Right support force will increase.d. Right support = 43 kNExplanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
Which material could be represented by the image above? (SC.8.P.8.1)
A. Diamonds
B. Milk
C. Oxygen
D. Wood
Answer:
a. diamonds
Explanation:
solid is the matter that has definite shape and definite volume. One good example is a diamond because contains particles that are strongly attract to each other.
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 7.63 m/s in 3.94 s. What is the magnitude of the linear impulse experienced by a 73.7 kg passenger in the car during this time? Submit Answer Tries 0/20 What is the average force experienced by the passenger?
Answer:
1. p = 562.3 kg*m/s
2. F = 142.7 N
Explanation:
1. The linear impulse (p) is given by:
[tex] p = mv [/tex]
Where:
m: is the passenger's mass = 73.7 kg
v: is the speed = 7.63 m/s
[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]
Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.
2. The average force can be calculated using the following equation:
[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]
Therefore, the average force experienced by the passenger is 142.7 N.
I hope it helps you!
Two identical items, object 1 and object 2, are dropped from the top of a 50.0m50.0m building. Object 1 is dropped with an initial velocity of 0m/s0m/s, while object 2 is thrown straight downward with an initial velocity of 13.0m/s13.0m/s. What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground
Answer:
Δt = 1.1 s
Explanation:
Given information:
H= 50.0 m
g= 9.8 m/s²
Object 1v₀ = 0[tex]H = \frac{1}{2} * g* t^{2}[/tex]
Solving for t, we get:
t₁= 3.2 s
Object 2v₀ = 13 m/sWe can find the final velocity for the object when it hits the ground, using the following expression:
[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]
Solving for vf, we get:
vf = 33.9 m/s
Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:
[tex]v_{f} = v_{o} + g*t[/tex]
(Assuming the downward direction to be positive).
Solving for t, we get:
t₂ = 2.1 s
So the difference in time when both objects hit the ground, it's simply
Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s
If vector A = 6i - 2j + 3k, determine
(a) A vector in the same direction as A with magnitude 2A
(b) A unit vector in the direction of A
(c) a vector opposite to A with magnitude of 4 m
Answer:
(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
Explanation:
Vectors
Given a vector
[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]
We must determine the following:
a) A vector in the same direction as A with double magnitude 2A.
If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:
[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
b) A unit vector in the same direction of A.
The unit vector needs to compute the magnitude of the vector:
[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]
[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]
[tex]\mid A\mid=7[/tex]
The unit vector is:
[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]
[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]
[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.
The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:
[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]
[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
The smokestack of a stationary toy tra in consists
of a vertical spring gun chat shoots a steel ball a meter
or so straight into the air-----so straight that the ball
always falls back into the smokescack. Suppose the
train moves at constant speed along the straight track. Do you think the ball will still return to the smoke-
stack if shot from the moving train? What if the train
gains speed along the straight track? What if it moves at a constant speed on a circular track? Discuss why your answers differ,
Answer:
i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
Explanation:
The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
correct me if im wrong
An object is dropped from rest and falls freely 20 m to Earth. When is the speed of the object 9.8 m/s?
At the end of the first second of its fall
At the end of the first second of its fall
During the entire time of its fall
During the entire time of its fall
At no time is the speed 9.8 m/s
At no time is the speed 9.8 m/s
During the entire first second of its fall
During the entire first second of its fall
After it has fallen 9.8 meters
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s
An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.
Answer:
6.7 seconds
Explanation:
d=(1/2)at^2
equation
1000=(1/2)45t^2.
substitute
2000=45t^2.
multiply by 2 for both sides
44.44=t^2.
divide both sides by 45
6.7=t
take the square root of both sides
A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and cresting a 2800 m pass. (a) Neglecting friction, what should the speed of the car be at the top of the second pass? (b) Find the actual speed of the car if the work due to nonconservative forces is – 5 x106 J.
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s
Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]
2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]
Explanation:
1. The strength of the nucleus' electric field (E):
[tex]E = \frac{kq}{r^{2}}[/tex]
Where:
k: is the Coulomb constant = 9x10⁹ Nm²/C²
q: is the proton charge = 1.6x10⁻¹⁹ C
r: is the radius = 10⁻¹⁰ m
[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]
2. The kinetic energy (Ek) of an electron is the following:
[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]
Where:
m is the electron's mass = 9.1x10⁻³¹ kg
v: is the speed of the electron
We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):
[tex] F_{c} = F_{e} [/tex]
[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]
[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]
Now, we can find the kinetic energy:
[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]
I hope it helps you!
how much power is used if it takes frank (a 450 N boy ) 3 seconds to run 2 meters ?
Answer:
300
Explanation:
450Newton × 2Meter ÷ 3sec
please answer this question with an explanation
will mark brainiest
Answer:
jenny
Explanation: