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Mandarinka [93]

3 years ago

10

A 0.5kg ball of clay originally moving at 6 m/s strikes a wall and comes to rest in 0.25s, what is the magnitude of the impulse

given to the ball of clay?
A) 0.75 kg m/s
B) 1.5 kg m/s
C) 3.0 kg m/s
D) 12 kg m/s

1 answer:

Alex787 [66]3 years ago

5 0

Answer:

C I did USA testprep

Explanation:

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A car travels around a curve. What can you say about the centripetal force if the velocity is tripled?

Lesechka [4]

Answer:

F = M a = M v^2 / R

If v is increased by three the force will be increased by nine,

C) is correct

4 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

miss Akunina [59]

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

(c) Particle C

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

3 0

3 years ago

Once broken into parts curved motion can be worked as ________________ problems along both axes.

vovikov84 [41]

Answer:

projectile motion

Explanation:

i am not sure sorry

8 0

4 years ago

PLEASE HELP !!!

emmainna [20.7K]

Answer and Explanation:

The answer is <u>D) Alpha and Gamma</u>

Gamma radiation does not cause transmutations.

A <u>transmutation</u> is the <u>conversion of an atom of one element to an atom of another. This generally occurs through nuclear reacting.</u>

There are three main types of radiation: alpha, beta, and gamma. Both beta and alpha decay cause changes to the mass and atomic numbers. This results in a transmutation. Gamma radiation, however, does not.

Gamma radiation is the result of a gamma ray. In essence, the nucleus emits a high-energy proton. This is very penetrating and can only be stopped by aluminum, lead, soil, water, and concrete. This type of radiation does not change the element and, therefore, does not cause a transmutation.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

3 0

3 years ago

One system consists of two like charges (q1 and q2) separated by a distance r. A second system consists of two charges that are

olganol [36]

fyt as per the question the magnitude of two like charges are given as q1 and q2.the separation distance is given as r unit.hence the potential energy is given as- $P.E=+\frac{1}{4\pi\epsilon} \frac{q1*q2}{r^2}$

here the potential energy is positive which means the force between two charges is repulsive.the potential energy is maximum which indirectly denotes that the system is unstable.due to this repulsion the smaller charge may accelerate.

coming to the same charges of opposite nature i.e unlike charges-here the magnitude of charges are same and separation distance is also are,so the potential energy will be given as- $P.E=-\frac{1}{4\pi\epsilon}\frac{q1*q2}{r^2}$

here the potential energy is negative .so the system of two charges are attracted by each other.

4 0

4 years ago

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