All categories

4 years ago

100 POINTS QUESTION. PLEASE PROVIDE EXPLANATION

Physics

2 answers:

melomori [17]4 years ago

4 0

Answer:

t = 0, p = 12.0, v = 6.00

t = 20, p = 112, v = 56.0

t = 25, p = 62.0, v = 31.0

Explanation:

Impulse = change in momentum

J = Δp

FΔt = mΔv

The impulse equals the area under the F vs t graph.

At t = 0 s, the initial velocity is 6.00 m/s, so the momentum is:

p = (2.00 kg) (6.00 m/s)

p = 12.0 kg m/s

From t=0 s to t=20 s, the impulse is:

J = (20 N) (5 s) + ½ (20 N) (5 s) + ½ (-10 N) (10 s)

J = 100 Ns

So the new momentum is:

p = 12.0 kg m/s + 100 kg m/s

p = 112 kg m/s

And the new velocity is:

v = (112 kg m/s) / (2.00 kg)

v = 56.0 m/s

From t=20 s to t=25 s, the impulse is:

J = (-10 N) (5 s)

J = -50 Ns

So the new momentum is:

p = 112 kg m/s − 50 kg m/s

p = 62.0 kg m/s

And the new velocity is:

v = (62.0 kg m/s) / (2.00 kg)

v = 31.0 m/s

diamong [38]4 years ago

4 0

Answer:

Hello!!! erz here ^^

Explanation:

First Section: t = 0s, p = 12.0, v = 6.00

Second Section: t = 20, p = 112, v = 56.0

Third Section: t = 25, p = 62.0, v = 31.0

Hope this helps!! :D

You might be interested in

In his famous experiment, Rutherford fired alpha particles at a thin gold film. Most of the alpha particles went through the fir

ale4655 [162]

Answer:

1. The nucleus is about 1/2 the size of the atom

Explanation:

Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size

Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.

Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles

so correct answer will be

1. The nucleus is about 1/2 the size of the atom

5 0

3 years ago

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

frozen [14]

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

3 0

4 years ago

Read 2 more answers

Why are viruses hard to fight

ki77a [65]

Answer:Compared to other pathogens, such as bacteria, viruses are minuscule. And because they have none of the hallmarks of living things — a metabolism or the ability to reproduce on their own, for example — they are harder to target with drugs.

Explanation:

7 0

3 years ago

What is formed when the rock making up a tectonic plate breaks and moves Select one of the options below as your answer:

dem82 [27]

A. a fault hopes this helps you guys

6 0

4 years ago

A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, th

erica [24]

QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.

6 0

3 years ago