Answer: 549.9 km
Explanation: 84.6km every hour so 84.6*6.5= 549.9
On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)
Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]
Therefore, the height of the cliff is 121.276 m
A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s. At this moment, the truck's momentum (in kg · m/s) is how many times as large as the truck's speed (in m/s)?IncorrectThis means that as the truck travels, the truck's momentum (in kg · m/s) is always Incorrect times as large as the truck's speed (in m/s).If the truck is traveling at 16 m/s, what is the truck's momentum?
Answer:
At the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
Explanation:
From the question,
The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).
Let the truck's momentum be P and the truck's speed be v,
Then we can write that
P∝v
Then,
P = kv
Where k is the proportionality constant
From the question,
At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,
To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is
50600 ÷ 23 = 2200
Hence, at the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
From
P = kv
Then, k = P/v
At a moment, P = 50600 kg · m/s and v = 23 m/s
∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg
k = 2200 kg
To determine the truck's momentum if the truck is traveling at 16 m/s
From
P = kv
k = 2200 kg
v = 16 m/s
∴ P = 2200 kg × 16 m/s
P = 35200 kg · m/s
Hence, if the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?
Answer:
At a distance of 100 m from the source the intensity will be 40 dB.
Explanation:
Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.
The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.
The conversion between intensity and decibels corresponds to:
[tex]L=10*log\frac{I}{I0}[/tex]
where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.
In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:
[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]
[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]
Being L1= 70 dB and L2= 40 dB
[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]
[tex]30=10*[log(\frac{I1}{I2})][/tex]
[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]
[tex]3=log(\frac{I1}{I2})[/tex]
[tex]10^{3} =\frac{I1}{I2}[/tex]
[tex]1,000=\frac{I1}{I2}[/tex]
The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:
[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]
Then:
[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]
Being d1= 10 m
[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]
[tex]1,000=\frac{d2^{2} }{100}[/tex]
1,000*100= d2²
10,000= d2²
√10,000= d2
100 m= d2
At a distance of 100 m from the source the intensity will be 40 dB.
A motor in a radio control car uses 7.2 volts and draws 14.4 Amps of electricity. What is the resistance of the motor?
Answer :
0.5 Ω
Explanation:
The following data were obtained from the question:
Voltage (V) = 7.2 V
Current (I) = 14.4 A
Resistance (R) =..?
The resistance of the motor can be obtained by applying the ohm's law equation as shown below:
V = IR
7.2 = 14.4 × R
Divide both side by 14.4
R = 7.2/14.4
R = 0.5 Ω
Therefore, the resistance is 0.5 Ω
Answer:
Answer :
0.5 Ω
Explanation:
The following data were obtained from the question:
Voltage (V) = 7.2 V
Current (I) = 14.4 A
Resistance (R) =..?
The resistance of the motor can be obtained by applying the ohm's law equation as shown below:
V = IR
7.2 = 14.4 × R
Divide both side by 14.4
R = 7.2/14.4
R = 0.5 Ω
Therefore, the resistance is 0.5 Ω
PLEASE SOLVE FAST!!! If the average American watches hours of TV every day , how many minutes will be spent in front of the TV by a person's 65th birthday? Solve using Dimensional Analysis.
Answer:
5694000 min
Explanation:
Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.
(4 h/day) × (60 min/1 h) = 240 min/day
They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.
240 min/day × (365 day/year) = 87600 min/year
They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.
87600 min/year × 65 year = 5694000 min
Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years
Answer:
T = 1.4696 10⁴ years
Explanation:
For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law
F = ma
G m M / r² = m a_c = m v² / r
G M / r = v²
the speed of the circular orbit is
v = 2π r / T
we substitute
G M / r = 4π² r² / T²
T² = (4π² / G M) r³
Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse
T² = (4π² /GM) a³
the constant is worth
(4π² / GM) = 2.97 10⁻¹⁹ s² / m³
let's reduce the distance to SI units
AU is the distance from the Earth to the Sun
a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)
a = 8.976 10¹³ m
T² = 2.97 10⁻¹⁹ (8.976 10¹³)³
T² = 21.4786 10²²
T = 4.63 10¹¹ s
Let's reduce to years
T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)
T = 1.4696 10⁴ years
A toy helicopter takes off and moves 6 m up and then 1 m back down. What
is the displacement of the helicopter?
A. 7 m
5 m up
C. 6 m
D. 1 m down
the answer is 5 m up AKA "B"
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Answer:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
An LED lamp powered by a USB-based portable battery has an effective resistance of 500 Ohm. If the battery is rated for 10,000 mAh, then how long can the lamp be powered and what is the total power consumed by the lamp? The operating voltage of a USB powered device is 5V. Assume that the battery is also rated at 5V.
Answer:
Time = 1000 h
Power = 0.05 W = 50 mW
Explanation:
First we will find the current consumed by the lamp. For this purpose we can use the Ohm's Law. The equation of Ohm's Law is written as follows:
V = IR
I = V/R
where,
I = Current = ?
V = Voltage = 5 V
R = Resistance = 500 Ω
Therefore,
I = 5 V/500 Ω
I = 0.01 A = 10 mA
Now the time duration of the operation f lamp can be found by:
Time = Rating/Current
Time = 10000 mAh/ 10 mA
Time = 1000 h
The power consumption of lamp is given as follows:
Power = IV
Power = (0.01 A)(5 V)
Power = 0.05 W = 50 mW
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
What are the standard international (si) units of distance
Answer:
meter
Explanation:
Answer: The International System of Units is a system of measurement based on 7 base units
Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.
In this circuit, all the resistors are identical, they are assigned voltage V1, V2 and V3 which correspond to R1 R2 and R3.
Put the volume in order of lowest to highest.
PLEASE HELP, WILL GIVE BRAINLIEST, NEED A GOOD EXPLANATION NOT JUST ANSWER AS IM VERY CONFUSED.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:
29 m/s2
31 m/s2
33 m/s2
30 m/s2
32 m/s2
Explanation:
The question is incomplete. Here is the complete question.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of 30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...
Given
initial velocity of rocket u = 0m/s
final velocity of rocket = 30m/s
Height reached by the rocket = 98m
Required
upward acceleration of the rocket
Using the equation of motion below to get the acceleration a:
[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]
Hence upward acceleration of the rocket during the burn phase is closest to 5m/s²
Note that the velocity used in calculation was assumed.
HELP PLS GIVING BRAINLIEST!!
A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .
Answer:
The magnitude is [tex]B = \frac{450}{7.9* 10^5}[/tex]
The direction is the positive x axis
Explanation:
From the question we are told that
The electric field is E = 450 V/m in the negative y ‑direction
The speed of the proton is [tex]v= 7.9* 10^5\ m/s[/tex] in the positive z direction
Generally the overall force acting on the proton is mathematical represented as
[tex]F_E = q(\vec E + \vec v * \vec B)[/tex]
Now for the beam of protons to continue un-deflected along its straight-line trajectory then [tex]F_E =0[/tex]
So
[tex] 0 = q( E (-y) + v(z) * \vec B)[/tex]
=> [tex]E\^y = v \^ z * \vec B[/tex]
Generally from unit vector cross product vector multiplication
[tex]\^ z \ * \ \^ x = \^ y[/tex]
So the direction of B (magnetic field must be in the positive x -axis )
So
[tex]E\^y = v \^ z * B\^ x [/tex]
=> [tex]E\^y = vB ( \^ z * \^ x) [/tex]
=> [tex]E\^y = vB ( \^y) [/tex]
=> [tex]E = vB [/tex]
=> [tex]B = \frac{450}{7.9* 10^5}[/tex]
=> [tex]B = 0.0005696 \ T [/tex]
What is the final velocity (in m/s) of a hoop that rolls without slipping down a 3.50-m-high hill, starting from rest
Answer:
8.29m/s
Explanation:
Given
Height of hill H = 3.50m
Initial velocity u = 0m/s
Required
Final velocity v
Using the equation of motion
v² = u²+2gH
v² = 0²+2(9.81)(3.5)
v² = 0+7(9.81)
v² = 68.67
v = √68.67
v = 8.29m/s
Hence the final velocity of the hoop is 8.29m/s
calculate sound intensity
Answer:
I dont know sorry,I hope you find what are you looking for.
in a controlled experiment do none of the variables change?
Answer:
Yes
Explanation:
The variables change in and experiment.
Answer:
If you are carefully enough to control everything, then everything that could change the result of your experiment won't happen.
Explanation:
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground
Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside the solenoid near its center
Answer:
The value is [tex]B = 0.0452 \ T [/tex]
Explanation:
From the question we are told that
The number of turns is N = 1000
The length is L = 50 cm = 0.50 m
The radius is r = 2.0 cm = 0.02 m
The current is I = 18.0 A
Generally the magnetic field is mathematically represented as
[tex]B = \mu_o * \frac{N }{L} * I[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0[/tex]
=> [tex]B = 0.0452 \ T [/tex]
The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The original voltage is [tex]V_o[/tex]
The new voltage is [tex]V =\frac{V_o}{2}[/tex]
The capacitance is [tex]C = 150\ nF = 150 *10^{-9} \ F[/tex]
The first resistance is [tex]R_i = 26 \Omega[/tex]
The second resistance is [tex]R_E = 200 \Omega[/tex]
Generally the equivalent resistance is
[tex]R_e = R_1 + R_E[/tex]
=> [tex]R_e = 26 +200 [/tex]
=> [tex]R_e = 226 \ \Omega [/tex]
Generally the time constant is mathematically represented as
[tex]\tau = RC[/tex]
=> [tex]\tau = 226 * 150 *10^{-9}[/tex]
=> [tex]\tau = 3.39 *10^{-5} \ s [/tex]
Generally the voltage is mathematically represented as
[tex]V = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]\frac{V_o}{2} = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]0.5 = e^{-\frac{t}{\tau} }[/tex]
=> [tex]ln(0.5) = {-\frac{t}{ 3.39 *10^{-5} } }[/tex]
=> [tex]ln(0.5) * 3.39 *10^{-5} = -t [/tex]
=> [tex]t = 2.35*10^{-5} \ s [/tex]
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is [tex] v = 350 \ m/s [/tex]
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by [tex]k = 60 cm = 0.6 \ m[/tex]
The frequency of both speakers is [tex]f = 700 \ Hz[/tex]
Generally the distance of the listener to the first speaker is mathematically represented as
[tex]L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]
[tex]L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]
[tex]L_1 = 5.39 \ m [/tex]
Generally the distance of the listener to second speaker at its new position is
[tex]L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]
[tex]L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]
[tex]L_2 = 5.64 \ m [/tex]
Generally the path difference between the speakers is mathematically represented as
[tex]pD = L_2 - L_1 = \frac{n * \lambda}{2}[/tex]
Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as
[tex]\lambda = \frac{v}{f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * \frac{v}{f}}{2}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=> [tex] 5.64 - 5.39 = \frac{1 * v}{2*700}[/tex]
=> [tex] v = 350 \ m/s [/tex]
The speed of sound in air is 350 m/s
Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,
L = √[(5 m)² + (4/2 m)²]
= √[25 m² + (2 m)²]
= √[25 m² + 4 m²]
= √29 m² = 5.39 m.
Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now
L' = √[(5 m)² + (4/2 + 0.6 m)²]
= √[25 m² + (2 m + 0.6 m)²]
= √[25 m² + (2.6 m)²]
= √[25 m² + 6.76 m²]
= √31.76 m²
= 5.64 m.
Now, the path difference when we first have destructive interference is
ΔL = L' - L
= 5.64 - 5.39
= 0.25
Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.
Now, since we have destructive interference for the first time, n = 0.
So, ΔL = (n + 1/2)λ
ΔL = (0 + 1/2)v/f
ΔL = v/2f
Making v subject of the formula, we have
v = 2fΔL
Substituting the values of the variables into the equation, we have
v = 2fΔL
v = 2 × 700 Hz × 0.25 m
v = 350 m/s
So, the speed of sound in air is 350 m/s
Learn more about interference of sound here:
https://brainly.com/question/1346741
Problem I Marcella (see warmup problem, above) gets her car moving steadily at 4m/s but suddenly someone stops ahead to assist her and parks their car 14 meters from the front of her car. Marcella grabs the car bumper and pulls very hard, with 200 N of force. The work she does transfers energy out, it reduces the K of the car, as it gradually approaches the other car. a) What is the initial kinetic energy before she tries to stop the car? b) What is the final kinetic energy, when her car hits the other car? What is the speed? c) Suppose the other person also slowed her car, pushing it from the front. How much force would be needed to stop her car 1 meter from the other car? [1 m allows the person not to be crushed!]
Answer:
Explanation:
a) KE = (1/2) * m * ([tex]v^{2}[/tex]) = F * d = 14m * 200N = 2800 m/N or 2.8 * [tex]10^{3}[/tex] m/N
b) 0J and 0m/s (if Marcella stopped after going 14 meters)
c) Known from part (a) that KE = 2800 J = F1 * d1,
2800J = F1 * (14m - 1m) => F1 = 2800J/13m = 215.384 N
which statement is correct about the strength of forces?
-Electrostatic forces are exactly 10 times stronger than gravitational forces.
-Electrostatic forces are exactly 10 times weaker than gravitational forces.
-Electrostatic forces are trillions of times stronger than gravitational forces.
-Electrostatic forces are trillions of times weaker than gravitational forces.
Answer:
Thanks!!!!! adding this so it doesn’t get deleted.
Explanation:
1. Electrostatic forces are trillions of times stronger than gravitational forces. 2. normal force and friction 3. contact forces 4. The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate. 5. The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.
The correct statement over here is that electrostatic forces are trillions of times stronger than gravitational forces. Hence, option C is correct.
What is an Electrostatic Force?One of the basic forces in the cosmos is electrostatic force. In the universe, there are four basic forces. These include gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Under the umbrella of electromagnetic force is electrostatic force. Two charges placed apart are subject to the electrostatic force. The size of each charged and the separation between them determines how much electrostatic force there will be.
When two charges of the same type are brought together, whether positive or negative, they repel one another. It is known as the electrostatic force of repelling when it operates among two charges that are similar.
Therefore, the electrostatic forces are trillions of times stronger than the gravitational forces.
To know more about Electrostatic Force:
https://brainly.com/question/9774180
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Calculate the force a 75 kg high jumper must exert in order to produce an acceleration that is 3.2 times the acceleration due to gravity.
Answer:
Explanation
According to Newton's second law of motion,
F = ma
m is the mass
a is the acceleration
If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g
The formula becomes;
F = m(3.2g)
F = 3.2mg
m= 75kg
g = 9.81m/s²
F = 3.2(75)(9.81)
F = 2,354.4N
Hence the force exerted by the jumper is 2,354.4N
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *
A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s
Answer:
D. 10 m/s
Explanation:
How can you tell if hair was forcibly removed?
Answer:Explanation:
A microscopic hair examination can also determine if a hair was forcibly removed, artificially treated or diseased. A comparison microscope can be used to compare a questioned hair to a known hair sample in order to determine if the hairs are similar and if they could have come from a common source.
The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point. You stand on a pier watching water waves and see 10.9 waves pass by in a time of 28 seconds.What is the period of the water waves? Round to nearest .01 and do not include units in your answer.
Answer:
T = 2.57 s
Explanation:
The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point.
No of waves observed = 10.9
Time taken, t = 28 s
We need to find the period of the water waves. Number of waves per unit time is called frequency. Let it is f. So,
[tex]f=\dfrac{n}{t}\\\\f=\dfrac{10.9}{28}\\\\f=0.389\ Hz[/tex]
If T is period of the wave. It is equal to the reciprocal of frequency.
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.389}\\\\T=2.57\ s[/tex]
So, the period of the water waves is 2.57 seconds.
When driving at slower speeds you need to use what type of steering
wheel movements compared to when driving at faster speeds? *
Answer:
slower speeds = larger and faster steering wheel movements
faster speeds = small and slow steering wheel movements
Explanation:
When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.
A student must determine the relationship between the inertial mass of an object, the net force exerted on the object, and the object’s acceleration. The student uses the following procedure. The object is known to have an inertial mass of 1.0kg .
Step 1: Place the object on a horizontal surface such that frictional forces can be considered to be negligible.
Step 2: Attach a force probe to the object.
Step 3: Hang a motion detector above the object so that the front of the motion detector is pointed toward the object and is perpendicular to the direction that the object can travel along the surface.
Step 4: Use the force probe to pull the object across the horizontal surface with a constant force as the force probe measures force exerted on the object. At the same time, use the motion detector to record the velocity of the object as a function of time.
Step 5: Repeat the experiment so that the object is pulled with a different constant force.
Can the student determine the relationship using this experimental procedure?
Answer choices:
A) Yes, because Newton’s second law of motion must be used to determine the acceleration of the object.
B) Yes, because the net force exerted on the object and its change in velocity per unit of time are measured.
C) No, because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed.
D) No, because knowing the net force exerted on the object and its change in velocity per unit of time is not sufficient to determine the relationship.
Answer:
C
Explanation:
In order to obtain data about the object’s velocity as a function of time, the object must move either toward or away from the motion detector.
The student cannot determine the relationship using this experiment ; ( C ) Because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed
Given that the aim of the experiment is to determine the relationship between mass of inertia of object, net force exerted and acceleration of the object we have to first obtain the object's velocity (i.e. Distance travelled by object / time ). and
To obtain velocity of object with respect to time, the object must move in either direction ( back or front ) from its parallel position away from the position of the detector ( motion )
Hence we can conclude that The student cannot determine the relationship using this experimental procedures.Because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed
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