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Hitman42 [59]
3 years ago
11

Calculate the acceleration of a bus whose speed changes from 7 m/s to 16 m/s over a period of 5 s.

Physics
1 answer:
Bezzdna [24]3 years ago
7 0
Let’s do this together!

Okay so the acceleration formula is vf-vi over time .

So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s

So the time is 5s so 9m/s divided into 5s is 1.8m/s/2

So the answer is 1.8m\s2
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A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much w
Alecsey [184]

Answer:

581 kJ, work was done by the system

Explanation:

According to the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released)

W is the work done by the system (positive if done by the system, negative if done by the surrounding)

In this problem,

Q=+757 kJ

\Delta U = +176 kJ

Therefore the work done by the system is

W=Q-\Delta U=757 kJ-176 kJ=+581 kJ

And the positive sign means the work is done BY the system.

5 0
3 years ago
which of the following is not a type of cosmic radiation? gamma-rays, x-rays, sound waves, or cosmic rays​
Archy [21]

Answer: sound waves or cosmic rays

Explanation:

Cosmic radiation consist of high energy particles,x-rays and gamma rays produce in space

7 0
3 years ago
Which phase of matter contains particles that split into ions and electrons?
Lemur [1.5K]

Answer:

Plasma

Explanation:

7 0
2 years ago
A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho
Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f

There is no kinetic energy in the initial state, nor potential energy in the end,

mgh+0=0+KE_f

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)

The inertia of the bodies is given by the equation,

I=\frac{m(R_1^2+R^2_2)}{2}

I=\frac{2(0.2^2+0.1^2)}{2}

I=0.05Kgm^2

On the other hand the angular velocity is given by

\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s

Replacing these values in the equation,

(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2

Solving for h,

h=2.86m

5 0
2 years ago
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

5 0
3 years ago
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