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satela [25.4K]
3 years ago
9

A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

reading of the car in km/hr?
Physics
1 answer:
777dan777 [17]3 years ago
4 0

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

K = \frac{1}{2}mv^{2}

By substituting the values

1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

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What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?
KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

6 0
3 years ago
Determine the frequency of light whose wavelength is 4.257 x10-7 cm
Bas_tet [7]
<span>Frequency x Wavelength = Speed of light Now, speed of light = 3 x 10^5 km/s = 3 x 10^8 m/s = 3 x 10^10 cm/s Frequency = speed/Wavelength = (3 x 10^10)/(4.257 x 10^-7) =7 x 10^16 Hz</span>
6 0
3 years ago
Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force F = 12.3 N

Displacement d = 10.2 m

Mass of rocket m = 0.663 Kg

From work energy theorem,

  W = \Delta K

 W_{t} - Wd - W_{g} = KE

Where W_{t} = thrust work W_{g} = gravitational work

KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2

KE = 59.2 -Wd

After cutoff kinetic energy is converted into potential energy,

KE = Wd' + mg\Delta h

Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

Work done by drag force is given by,

Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

5 0
3 years ago
A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro
Anastaziya [24]

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.

Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12

Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

7 0
3 years ago
your friend sit in a sked in the snow. if you apply a force of 120 N to them, they have an acceleration of 1.3 m/s2. what is the
levacccp [35]
Need to know the equation for force
F=MA
F is force
M is mass- we need to know the mass
A is acceleration
use "x" for mass
120 N= 1.3x
divide 1.3 in both side
kg unit for mass
X=92.31 kg 
or 
mass = 92.31 kg
Hope this helps
7 0
4 years ago
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