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3 years ago

5

A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42

m/s?
250 s
0.14s
7.0 s
290 s

1 answer:

Over [174]3 years ago

4 0

There are few equations that link acceleration and velocity.

One of them is derived from the definition of acceleration - is the rate of change of velocity in respect with time

a = v-u/t
at=v-u

a = acceleration
v=velocity at the end
u=velocity at the start
t=time

Now put everything you know and solve for time.

6t = 42 -0
6t = 42
t = 42 / 6
t = 7 second

It takes 7 seconds for the roccet to reach a velocity of 42 m/s

Hope it helps :).

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A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have

$\sum F = 205.9 N$ is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2$

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

$a=1.37 m/s^2$

s = 350 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s$

4)

For this part of the problem, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

$a=1.37 m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s$

Learn more about Newton's laws of motion and accelerated motion:

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6 0

3 years ago

I need some help w/ this:

pochemuha

Answer:

The velocity of the police car is, v = 17.798 m/s

Explanation:

Given data,

The actual frequency of the siren, f = 2010 Hz

The observed frequency of siren is, f' = 2120 Hz

The velocity of the observer, v' = 0 m/s

The velocity of the source, v = ?

The formula for Doppler effect,

$f'=\frac{(V+v')}{(V-v)}f$

Where,

V - velocity of sound waves in air.

$v=V-(V+v')\frac{f}{f'}$

Substituting the given values,

$v=343-(343+0)\frac{2010}{2120}$

v = 17.798 m/s

Hence, the velocity of the police car is, v = 17.798 m/s

5 0

3 years ago

A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^

Nataliya [291]

Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2 m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low

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3 years ago

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A sled with mass 8.00kg moves in a straight line on a frictionless horizontal surface. At one point in it's path, it's speed is

kati45 [8]

We will use two definitions to solve this problem. The first will be given by the conservation of energy, whereby the change in kinetic energy must be equivalent to work. At the same time, work can be defined as the product between the force by the distance traveled. By matching these two expressions and clearing for the Force we can find the desired variable.

$W = KE_f-KE_i$

$Fd = \frac{1}{2}mv_f^2-\frac{1}{2} mv_i^2$

Thus the force acting on the sled is,

$F = \frac{m}{2s} (v_f^2-v_i^2)$

Replacing,

$F = \frac{8}{2(2.5)}(6^2-4^2)$

$F = 32N$

Therefore the Force acting on the sled is 32N

8 0

3 years ago

Albert is piloting his spaceship, heading east with a speed of 0.92 cc relative to Earth. Albert's ship sends a light beam in th

ohaa [14]

Answer:

He sees the light as 1c

Explanation:

According to relativity, the speed of light is the same in all inertial frame of reference.

If we were to add the velocities as applicable to a normal moving bodies, the relative speed of the light beam will exceed c which will break relativistic law since nothing can go past the speed of light.

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3 years ago