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3 years ago

John pushes his brother Danny on a skateboard. John applies a force of 29 N and Danny's acceleration is 0.4 m/s

Physics

2 answers:

belka [17]3 years ago

4 0

Answer:

Explanation:

Mumz [18]3 years ago

4 0

Answer:

The combined mass of Danny and his skateboard = m = 72.5 kg

Explanation:

Force applied by john = F = 29N

Danny’s Acceleration = a = 0.4 m/s

Combined mass of Danny and his skateboard = m = ?

According to Newton’s second law of motion:

F = ma

m = F/a

m = 29/0.4

m = 72.5 kg

Hence, the combined mass of Danny and his skateboard will be 72.5 Kg.

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PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball

bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

q₁ = 3C

q₂ = 14C

Distance between balls = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

F = $\frac{k q_{1} q_{2} }{r^{2} }$

where k = 9 x 10⁹Nm²/C²

q is the charges

r is the distance

Input the variables and solve;

F = $\frac{9 x 10^{9} x 3 x 14 }{9000}$ = 4.2 x 10⁷N

8 0

3 years ago

At an amusement park, a 7.00 kg swimmer uses a water slide to enter the main pool. The swimmer starts at rest, slides without fr

telo118 [61]

Answer:

27.44 J

Explanation:

We can find the energy at the top of the slide by using the potential energy equation:

PE = mgh

At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.

The swimmer's mass is given as 7.00 kg.

The acceleration due to gravity is 9.8 m/s².

The (vertical) height of the water slide is 0.40 m.

Substitute these values into the potential energy equation:

PE = (7.00)(9.8)(0.40)
PE = 27.44

Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.

Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.

6 0

3 years ago

A rocket blasts off from the Earth's surface. During the initial phase of flight, the engine of the rocket burns fuel at a rate

Katen [24]

Answer:

αβ = Ma

Explanation:

By Newton's 2nd Law, the equation governing the motion of the rocket while the rocket is burning fuel is

αβ = Ma where α = rocket's fuel burning rate, β = relative to the velocity of the rocket, M = instantaneous mass of the rocket and a = acceleration of rocket.

5 0

3 years ago

4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c

Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

6 0

3 years ago

A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Arturiano [62]

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

time taken for the acceleration, $t=5.73\ s$

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$

5 0

3 years ago