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2 years ago

A zebra runs across a field at a constant speed of 14m/s how far does the zebra go in 8 seconds?

Physics

2 answers:

Ratling [72]2 years ago

3 0

Answer:

112m/s

Explanation:

14x8=112 therefore meaning the zebra would run 112m/s

RoseWind [281]2 years ago

3 0

A zebra runs across a field at a constant speed of 14 m/s. The zebra would run 112 m in 8 seconds.

Explanation: 

According to Kinematics, the distance travelled is given by the product of speed and time. Here, the speed given as 14 m/s, and time is 8 seconds. Now, the equation is as follows,

$\text {velocity} = \frac{\text {distance}(s)}{\text {time}(t)}$

The required term to find as per the question is distance, so rearranging the above equation, we get,

$s = v e l o c i t y \times t$

By substituting the given values, we get, how much distance the zebra can run in time period of 8 seconds as,

Distance, $s = 14 \times 8 = 112 m$

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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

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2 years ago

The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b

Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

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2 years ago

The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1

antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

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2 years ago

Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

frosja888 [35]

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where v is the wave's speed and $\lambda$ is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

$f=\frac{400 m/s}{2 m}=200 Hz$

- underwater, the frequency of the wave is:

$f=\frac{1600 m/s}{8 m}=200 Hz$

So, the frequency has not changed.

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2 years ago

Read 2 more answers

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end tex

love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

$\Delta x = v_{i}.t + \frac{a}{2}.t^{2}$