What is the frequency of a wave that has a period of 5 seconds?
O A. 2 Hz
OB. 5 HZ
O C. 20 Hz
O D. 0.2 Hz
Answer: D
Explanation:
The frequency of a wave that has a period of 5 seconds is 0.2 Hz. Thus, the correct option is D.
What is Frequency?Frequency of a wave is the number of occurrences of a repeating event per unit of the time. Frequency is also occasionally referred to as the temporal frequency for clarity purpose, and it is distinct from that of angular frequency. Frequency is measured in SI units of hertz which is equal to one event per second of the time.
Frequency can be calculated as the number of waves that pass a fixed place in a given amount of time period. The frequency of a wave that has a period of 5 seconds is:
f = 1/T
where, f = frequency,
T = Time taken
f = 1/ 5
f = 0.2 hertz
Thus, the frequency of the wave is 0.2 hertz.
Therefore, the correct option is D.
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The zygote eventually becomes a(n)
sperm
egg
baby
uterus
HURRY 30 MINUTES LEFT
(SCIENCE )
Answer:
baby
Explanation:
Answer:
The zygote eventually becomes a baby
Explanation:
Got it right on a unit test in edg 2020
A wave has a frequency of 8 Hz, how long is the period?
A 16
B .125
C 64
D 8
Answer:
A is the answer because if you multiply 8 and 2 the period of wave frequency is 16Hz long.
Explanation:
Answer:
B 0.125 its a simple conversion
Question #005- List 10 active things you can do with your family/parents this weekend? (Please be sure to number them.)
i suck at this lol
Cycling
Walking
Shopping
Exercising
Dancing
Playing physical games
Sport
Cooking
Working around the house
Gardening
Hope this helps:D
口
Fgrav = 600 N
I need help
If a car that is moving 20.0 m/s has a momentum of 29000 kg·m/s, what mass is the car?
Answer:
Mass = 1450kg
Explanation:
P = M * V (where p is momentum, m is mass and v is velocity)
29000 = 20 * M
M = 29000 / 20
M= 1450 kg
At the start of a hockey game the record a job at the puck between two players from the opposing team. Each player wants to push the puck in the opposite direction. For several seconds the park does not move even though both players are pushing on it with the hockey sticks. What forces are acting on the puck, why the park does not move, and how one of the players could get the park to move in a direction they want
Answer:
1) The forces acting on the puck are the forces applied by the two players in their bid to push the puck in the opposite direction
2) The puck does not move because the forces applied by the two players are equal and opposite
3) A player can get the puck to move in the direction they want by increasing the force acting in the direction they want the puck to move
Explanation:
Two conducting loops of the same diameter, #1 made of copper and #2 made of aluminum wires of the same length and cross-sections are placed on the horizontal surface in uniform magnetic field directed vertically up. Magnetic field vanishes suddenly. The magnitude of emf in loop #1 is
Answer:
Same as loop #2
Explanation:
The magnitude of emf in loop #1 is the same as that of loop #2
A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.
Explanation:
the weight of the people inside the bus
The v
h.
The velocity of a vehicle reaches 4 m/s in 4 seconds. Is the velocity of the
car uniform? Explain.
Answer:
The velocity is not uniform
Explanation:
The velocity is not uniform, since the vehicle is increasing its speed to a value of 4 [m/s] in a time of 4 [s], that is, the vehicle is accelerating to increase its velocity, so when there is acceleration the velocity can not be constant or uniform.
Vf = Vo +(a*t)
where:
Vf = final velocity = 4 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
t = time = 4 [s]
a = (4 - 0)/4
a = 1 [m/s²]
if an object measures to be 80 centimeters in length, then it is equivalent to ____
Answer:
31.496 inches
Explanation:
Please do This:
I will name the brainliest!!!
Answer:
hi
Explanation:
A physics student tossed a ball vertically straight upward.
Describe the magnitude of the ball’s vertical velocity at its highest position.
a. The magnitude of vertical velocity is zero.
b. The magnitude of vertical velocity is 9.8 m/s.
c. The magnitude of vertical velocity is at a maximum value.
d. The magnitude of vertical velocity cannot be determined.
Answer:
c. The magnitude of vertical velocity is at a maximum value.
Explanation:
I think this is correct I'm not entirely sure though
Answer:
Explanation: ns along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. (Figure) illustrates the notation for displacement, where \mathbf{s} is defined to be the total displacement and \mathbf{x} and \mathbf{y} are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation \mathbf{A} to represent a vector with components {\mathbf{A}}_{x} and {\mathbf{A}}_{y}. If we continued this format, we would call displacement \mathbf{s} with components {\mathbf{s}}_{x} and {\mathbf{s}}_{y}. However, to simplify the notation, we will simply represent the component vectors as \mathbf{x} and \mathbf{y}.)
Review of Kinematic Equations (constant a)
x={x}_{0}+\stackrel{-}{v}t
\stackrel{-}{v}=\frac{{v}_{0}+v}{2}
v={v}_{0}+\text{at}
x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}
{v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\text{.}
The total displacement \mathbf{s} of a soccer ball at a point along its path. The vector \mathbf{s} has components \mathbf{x} and \mathbf{y} along the horizontal and vertical axes. Its magnitude is s, and it makes an angle \theta with the horizontal.
A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.
Given these assumptions, the following steps are then used to analyze projectile motion:
Step 1.Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so {A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta and {A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta are used. The magnitude of the components of displacement \mathbf{s} along these axes are x and \mathrm{y.} The magnitudes of the components of the velocity \mathbf{v} are {v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta and {v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta ,} where v is the magnitude of the velocity and \theta is its direction, as shown in (Figure). Initial values are denoted with a subscript 0, as usual.
Step 2.Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:
\text{Horizontal Motion}\left({a}_{x}=0\right)
x={x}_{0}+{v}_{x}t
{v}_{x}={v}_{0x}={v}_{x}=\text{velocity is a constant.}
\text{Vertical Motion}\left(\text{assuming positive is up}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9.\text{80}{\text{m/s}}^{2}\right)
y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t
{v}_{y}={v}_{0y}-\text{gt}
y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\mathrm{gt}}^{2}
{v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right)\text{.}
Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.
Step 4.Recombine the two motions to find the total displacement\mathbf{\text{s}} and velocity \mathbf{\text{v}}. Because the x – and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}} and \theta ={\text{tan}}^{-1}\left({A}_{y}/{A}_{x}\right) in the following form, where \theta is the direction of the displacement \mathbf{s} and {\theta }_{v} is the direction of the velocity \mathbf{v}:
Total displacement and velocity
s=\sqrt{{x}^{2}+{y}^{2}}
\theta ={\text{tan}}^{-1}\left(y/x\right)
v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}
{\theta }_{v}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)\text{.}
(a)
Two technicians are discussing the FMVSS 135 standards for parking brakes. Technician A states that the hand force required to set the brake should not exceed 80 lb. Technician B states that the foot force to set the brake should not exceed 80 lb. Which technician is correct?
Answer:
Technician A
Explanation:
Technician A is correct because, the hand force required is not supposed and even should not exceed 80 lb.
On the other hand, the foot force limit is even higher at 100 lb. Thus, the required foot force break should not exceed 100 lb. We are given 80 lb in the question, which is quite less than 100 lb. It could exceed 80, but must not exceed 100. Thus only Technician A is correct among them both.
Click The Pros and Cons of Plastics and use the information to answer the question.
Which scenario shows an indirect environmental impact of using plastics?
Answer:
Plastics being consumed by fish and passing toxic compounds to the humans eat the fish.
Explanation:
I did that answer and got it right.
Answer:
Plastics being consumed by fish and passing toxic compounds to the humans that eat the fish.
Explanation:
What is the density of this liquid that has a mass of 300g and volume 600ml.
Answer:
0.5
Explanation:
D = M / V
You divide 300g by 600ml and you get 0.5
(✿◠‿◠)
la respuesta es
0.5
espero que te sirva
:-)
A 52-kg bike is moving along a smooth road at a constant velocity of 8.4 m/s. What is the net force acting on the bike
Answer:
0 N
Explanation:
Net force = mass × acceleration
∑F = ma
The velocity is constant, so the acceleration is 0. Therefore, the net force is 0.
The net force acting on the bike is 0 N.
What is net force?Net force is the sum of all forces acting on an object. The net force can be calculated using Newton's second law, which states that F = ma, where: F is the net force. m is the mass of the object. a is acceleration.
The velocity is constant, so the acceleration is 0. Therefore, the net force is 0 N.
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can someone help me answer this? i dont get it and I need help before the next 30 minutes. ill give a brainiest to whoever helps me tyyy :)
Answer:
S=5.83...
Explanation:
s=d/t
Speed=distance/time
s=35/10
s=5.83...
A fixed particle with charge –8.8 nC and a second particle with charge –4.3 nC, respectively, are initially separated by a distance of 0.11 m. They are released and the second particle moves 0.030 m. A) What is the change in electric potential energy? B) Did the second particle move toward or away from the source charge?
Answer:
(a). The change in electric potential energy is 8.3 μJ.
(b). The second particle moves away from the source charge.
Explanation:
Given that,
Charge of first particle = -8.8 nC
Charge of second particle = -4.3 nC
Distance = 0.11 m
They are released and the second particle moves 0.030 m,
(a). We need to calculate the change in electric potential energy
Using formula of potential energy
[tex]U=\dfrac{Kq_{1}q_{2}}{d}[/tex]
Change in potential energy
[tex]\Delta U=\dfrac{kq_{1}q_{2}}{d_{2}}-\dfrac{Kq_{1}q_{2}}{d_{1}}[/tex]
Put the value into the formula
[tex]U=\dfrac{9\times10^{9}\times(8.8\times10^{-9}\times4.3\times10^{-9})}{0.030}-\dfrac{9\times10^{9}\times(8.8\times10^{-9}\times4.3\times10^{-9})}{0.11}[/tex]
[tex]\Delta U=0.000008256\ J[/tex]
[tex]\Delta U=8.3\ \mu J[/tex]
(B). We need to find the second particle move toward or away from the source charge
We know that,
Both charges are same, so the second particle will be repul from the source charge.
So, The second particle moves away from the source charge.
Hence, (a). The change in electric potential energy is 8.3 μJ.
(b). The second particle moves away from the source charge.
an someone help me please, thanks.
A rock is dropped off a cliff and strikes the ground with an impact velocity of 20m/s. How high was the cliff?(use a=-10m/s2)
Answer:
20 m
Explanation:
by the formula,
[tex]2aS = V_{f}^{2} - V_{i}^2[/tex]
2 x 10 x S = 20^2 - 0
20S = 400
[tex]S = \frac{400}{20}[/tex]
S = 20 m
A physics book is moved once around the perimeter of a table of dimensions 1.0m by 2.0 m. a. if the book ends up in its initial position, what is its displacement? Type here b. What is the distance traveled?
Answer:
A. 0 m.
B. 6 m.
Explanation:
Displacement is final position to initial position in the shortest length possible, if you end at your initial position then your displacement is basically 0.
Since the book traveled along the perimeter of a table of 1m and 2m then its perimeter is a total of 6m, which is basically how far the book traveled.
what of the average speed from 4s from 8s
A.0.5m/s
B.1m/s
C.2m/s
D.3m/s
E.9m
=====================================================
Explanation:
At 4 seconds, the distance is 8 meters. Note the point (4,8)
At 8 seconds, the distance is 10 meters. The point here is (8,10)
Find the slope of the line through (4,8) and (8,10)
m = (y2-y1)/(x2-x1)
m = (10-8)/(8-4)
m = 2/4
m = 1/2
m = 0.5
The average speed on this interval is 0.5 m/s
-----------------
Another way to look at it:
We've gone from 8 meters to 10 meters, so that's a change of y = 2 meters.
During this, the time has changed from 4 seconds to 8 seconds, which is a difference of x = 4 seconds.
The velocity is the rate of change of distance over time
velocity = (change in distance)/(change in time) = y/x = 2/4 = 0.5 m/s
The diagram represents Earth's orbit around the Sun. Where is the Sun located in this ellipse?
A)
At the center
B)
At the periphery
At the epicenter
D)
At one of the two foci
Submit
Helpppppp
Answer:
Its a D.
Explanation:
A car has a velocity of 10m/s.it now accelerates for 1m/s for 1/4 minutes. Find the distance travelled in this time and final speed of the car
¡Hellow!
For this problem, lets recabe information:
v (Velocity) = 10 m/s
a (Aceleration) = 1 m/s²
t (Time) = 1/4 min = 25 s
d (Distance) = ?
v' (Final velocity) = ?
First, for calculate distance, lets applicate formula:
[tex]\boxed{\boxed{\text{d = Vo * t + (a * t}^{2})\text{ * 0,5} } }[/tex]
Lets replace according we information and let's resolve it:
d = 10 m/s * 25 s + (1 m/s² * (25 s)²) * 0,5
d = 250 m + (625 m) * 0,5
d = 2,5 m + 312,5 m
d = 314 meters.
Now, for calculate final speed, lets applicate formula:
[tex]\boxed{\boxed{\text{v' = v + a * t} } }[/tex]
Lets replace according we information and let's resolve it:
v' = 10 m/s + 1 m/s² * 25 s
v' = 10 m/s + 25 m/s
v' = 35 m/s
¿Good Luck?
Att: That guy who use the "ñ".
A car decelerates uniformly from 24 m/s to 18 m/s in 3.0 seconds. How far does it travel during this time?
Distance travelled will be = 63 m
What is equation of motion?
These are the equations that describe the behavior of a physical system in terms of its motion as a function of time .
Given : u = 24 m/s
v= 18 m/s
t = 3sec
using equation of motion : v = u + at
18 = 24 + a*3
-6 = 3a
a = -2 m/s^2
using equation of motion : s = u t + 1/2 a (t^2)
s = 24 * 3 + 1/2 * (-2)(9)
s = 63 m
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Urgent and easy
Give an example where the following energy changes would take place:
7) Electrical to Thermal-
8) Chemical to Thermal-
9) Electrical to Mechanical-
10) Radiant to Chemical-
Answer:
7) Oven, toaster. 8) Fire, gas stove. 9) Television? No.. umm.. generator, yeah a generator 10) photosynthesis
Explanation:
Three identical 6.4kg masses are hung by three identical springs. Each spring has a force constant of 7.8 Kn/m and is 12cm long before any masses are attached to it. how long is the bottom most spring going to be after the three masses are hung on it
a. 14.3 b. 16.2 c.12.8 d.10.7
Answer:
The correct option is;
c. 12.8 cm
Explanation:
The given parameters are;
The length of each spring = 12 m = 0.12 m
Given that the masses and the springs are vertically oriented, we have;
The mass on the first spring = 3 × 6.4 = 19.2 kg
The weight on the first spring = 19.2 kg × 9.81 m/s² = 188.352 N
The extension of the first spring = 188.352 N/7,800 N/m ≈ 0.02415 m
The mass on the second spring = 2 × 6.4 = 12.8 kg
The weight on the second spring = 12.8 kg × 9.81 m/s² = 125.568 N
The extension of the second spring = 125.568 N/7,800 N/m ≈ 0.016098 m
The mass on the third spring = 1 × 6.4 = 6.4 kg
The weight on the third spring = 6.4 kg × 9.81 m/s² = 62.784 N
The extension of the third (bottom) spring = 62.784 N/7,800 N/m ≈ 0.00805 m
The total length of the bottom spring = Original length of the bottom spring + Extension of the bottom spring
The total length of the bottom spring ≈ 0.12 m + 0.00805 m = 0.12805 m
The total length of the bottom spring ≈ 0.12805 m ≈ 12.81 cm
The bottom spring will be approximately 12.81 cm long.
Why do you think the lunar cycle is longer than the time it takes the moon to orbit Earth?
Answer:
because the moon returns to the same place on the sky once every siderial period, but the sun is also moving on the sky.
Explanation:
Hope that helps!
Answer:
Because the moon returns to the same place on the sky once every siderial period (27.3 days), but the sun is also moving on the sky.Cycle of lunar phases takes 29.5 days. This is called the SYNODIC PERIOD
Explanation:
On Monday, Sylvester walked up 50 stairs in 1 minute. On Tuesday, Sylvester walked up 100 stairs in 2 minutes. Which best describes Sylvester's work and power? a.He did equal work and used equal power on both days. b.He did equal work both days, but used more power on Tuesday. c.He did more work on Tuesday, and used more power on Tuesday. d.He did more work on Tuesday, but used equal power on both days.