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Eduardwww [97]
3 years ago
9

An object in circular motion has velocity that is constantly changing. The direction of the acceleration is

Physics
1 answer:
Keith_Richards [23]3 years ago
7 0
<h2>Answer: Toward the center of the circle.</h2>

This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.  

However, in this movement the velocity has a constant magnitude, but its direction varies continuously.

Let's say \vec{V} is the velocity vector, whose direction is perpendicular to the radius r of the trajectory, therefore   the acceleration \vec{a} is directed toward the center of the circumference.

 

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If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d
azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt     from this equation we clear "is" = fs - gt

Substitution                         is = 77 - (9,81)(6.5)

Process                               is = 77 - 63.8

                                            is = 13.2 m/s

8 0
3 years ago
The radius of the planent venus is nearly the same as that of the earth,but its mass is only eighty percent that of the earth. I
oee [108]
Weight=mg
g=GM/r^2
g on venus is 0.80w as radius is kept constant
m of object is kept constant
w α g
w(venus( is 0.8w

7 0
3 years ago
Read 2 more answers
A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion
Tcecarenko [31]

Answer: 0.47 rad/sec

Explanation:

By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:

ω = ∆θ/ ∆t

Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:

ω = 2 π / 13.3 rad/sec = 0.47 rad/sec

6 0
3 years ago
Read 2 more answers
A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
Evgen [1.6K]

Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

6 0
3 years ago
Read 2 more answers
At what speed, as a fraction of c , is a particle's total energy twice its rest energy
WINSTONCH [101]
The rest energy of a particle is
E_0=m_0 c^2
where m_0 is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
E=mc^2 =  \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
E=2E_0
which means:
\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2
\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2
From which we find the ratio between the speed of the particle v and the speed of light c:
\frac{v}{c}=  \sqrt{1- (\frac{1}{2})^2 }  =0.87
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
3 0
3 years ago
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