All categories

3 years ago

calculate the time required for a 6000.-Newton net force to stop a 1200.-kilogram car initially traveling at 10. M/s

1 answer:

7 0

   Force = (mass) x (acceleration).

We know the force and the mass, so we can write

   6,000 N = (1,200 kg) x (acceleration).

Divide each side by 1,200 kg:

   Acceleration = 6000N/1200kg = 5 m/s² .

That's the acceleration. The car's speed changes by 5 m/s each second
that the force acts on it. If the force pushes from behind, then the car goes
5 m/s faster every second. If the force pushes from in front, then the car
goes 5 m/s slower every second.

The car is moving at 10 m/s and we want to slow it down to zero m/s. So
the force has to push from the front, and it will take (10/5) = <em>2 seconds</em> to
complete the job.

You might be interested in

Potential energy increases with the greater height mass of an object.True or False

guapka [62]

Answer:

Potential energy increases with the greater height mass of an object.

5 0

4 years ago

The diagram shows a wave. The quantity shown is the amplitude. This quantity is directly proportional to which of these other qu

Katen [24]

As we know that frequency and wavelength are dependent on each other

and this is given by

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

here we know that

$\lambda$ = wavelength

f = frequency

c = speed of wave

while for energy and intensity of wave we know that

Energy = $KA^2$

here A = amplitude of wave

so energy directly depends on amplitude of wave

so correct answer will be

<em>A) Energy</em>

3 0

4 years ago

Read 2 more answers

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Two balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L

olga_2 [115]

Answer:

Explanation:

Given

mass of balls $m= 5 kg$

$N=45.6 rev/s$

angular velocity $\omega =2\pi N=286.55 rad/s$

Length of Rod $2L=1.1 m$

Tension in the Second half of rod

$T_2=m\omega ^2(2L)=2m\omega ^2L$

$T_2=5\times (286.55)^2\times 1.1$

$T_2=451.609 kN$

For First Part

$T_1-T_2=m\omega ^2L$

$T_1=T_2+m\omega ^2L$

$T_1=3 m\omega ^2L$

$T_1=3\times 5\times (286.55)^2\times 0.55$

$T_1=677.41 kN$

7 0

3 years ago

Please help me with this physics question!

Vadim26 [7]

Not sure about the answer

7 0

4 years ago