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Romashka [77]
2 years ago
8

On one stretch of road, any vehicle travelling faster than 25 m/s is breaking the speed limit. The detectors are placed 1.2 m ap

art. Calculate the speed of a car that takes 0.050 s to travel this distance. Is it breaking the speed limit?​
Physics
1 answer:
Nadya [2.5K]2 years ago
4 0

Answer:

Speed = 24 m/s.

No! It doesn't break the speed limit.

Explanation:

The formula we'll be using to find the car's speed:

\boxed{ \mathsf{speed =  \frac{distance}{time} }}

The detectors are placed 1.2 m apart.

==> Distance the car travels = 1.2 m

The car, it says, takes 0.050 s to travel that distance.

==> Time taken by the car to travel 1.2 m = 0.050 s

Using the formula mentioned above, let's calculate the speed of the car:

\sf{speed =  \frac{distance}{time} }

\implies \sf{speed =  \frac{1.2}{0.050} }

<em>On</em><em> </em><em>dividing</em><em> </em><em>1</em><em>2</em><em> </em><em>by</em><em> </em><em>0</em><em>.</em><em>0</em><em>5</em><em>0</em><em> </em><em>we</em><em> </em><em>obtain</em><em> </em><em>2</em><em>4</em><em>.</em>

\implies \sf{speed = 24 \: m/s }

Any car traveling with a speed that is greater than or equal to <u>2</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u> is breaking the speed limit.

And since the car in our focus is traveling at 24 m/s that is less than 25 m/s, it doesn't break the speed limit.

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exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the
mash [69]

Answer:

\Delta x = v_0 t + \frac{1}{2}at^2

Explanation:

To solve this problem, we can use the following suvat equation:

\Delta x = v_0 t + \frac{1}{2}at^2

where

\Delta x is the vertical displacement of the frog

v_0 is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from v_0, all the other quantities are known. In fact:

\Delta x =0.1 m is the vertical displacement

t = 2 s is the total time of flight

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

Therefore, solving for v_0, we find the initial velocity of the frog:

v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s

4 0
3 years ago
A square loop of wire is held in a uniform 0.24 T magnetic field directed perpendicular to the plane of the loop. The length of
NNADVOKAT [17]

Answer:

Explanation:

Given that,

Magnetic field of 0.24T

B = 0.24T

Field perpendicular to plane i.e 90°

Rate of decrease of length of side of square is 5.4cm/s

dL/dt = 5.4cm/s = 0.054m/s

Since it is decreasing

Then, dL/dt = -0.054m/s

When L is 14cm, what is the EMF induced?

L = 14cm = 0.14m

EMF is give as

ε = - dΦ/dt

Where flux is given as

Φ = BA

Where A is the area of the square

A = L²

Then, Φ = BL²

Substituting this into the EMF

ε = - dΦ/dt

ε = - d(BL²)/dt

B is constant

ε = - Bd(L²)/dt

ε = -2BL dL/dr

ε = -2 × 0.24 × 0.14 × -0.054

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8 0
3 years ago
A diffraction grating with 1000 lines per mm is used in a spectrometer to measure the wavelengths of light emitted from a gas di
frez [133]

Answer:

= 9.8°

Explanation:

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width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m

angular position of fringe, Sinθ  = n λ /a

n is order of fringe , λ is wave length of light and a  is slit aperture

So Sinθ  ∝ 1 / a

Sin θ₁ /Sin θ₂ = a₂/a₁ ;

Sin20°/sinθ₂ = 2 / 1

sinθ₂ = Sin 20° / 2 = .342/2 = .171

θ₂ = 9.8 °

4 0
3 years ago
A crate slides down a ramp that makes a 20o angle with the ground. To keep the crate moving at a steady speed, Paige pushes back
cupoosta [38]

Answer:

-223.64684 J

Explanation:

F = Force that is applied to the crate = 68 N

s = Displacement of the crate = 3.5 m

\theta = Angle between the force and displacement vector = (180-20)

Work done is given by

W=Fscos\theta\\\Rightarrow W=68\times 3.5\times cos(180-20)\\\Rightarrow W=-223.64684\ J

The work that Paige does on the crate is -223.64684 J

6 0
2 years ago
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