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harina [27]
3 years ago
15

Velocity is described by __ and time

Physics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

direction and time

Explanation:

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a
____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, v=2.79\times 10^8\ m/s

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s

Time period of oscillation measured by the observer is :

t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0
3 years ago
The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

8 0
3 years ago
Read 2 more answers
An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
3 years ago
A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
muminat

Answer:

The speed of the plank relative to the ice is:

v_{p}=-0.33\: m/s

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

m_{g}v_{g}+m_{p}v_{p}=0 (1)

Where:

  • m(g) is the mass of the girl
  • m(p) is the mass of the plank
  • v(g) is the speed of the girl
  • v(p) is the speed of the plank

Now, as we have relative velocities, we have:

v_{g/b}=v_{g}-v_{p}=1.55 \: m/s (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

45v_{g}+168v_{p}=0

v_{g}-v_{p}=1.55

v_{p}=-0.33\: m/s

I hope it helps you!      

8 0
3 years ago
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

And velocity after rebound v_2=-19m/sec

We have to find the change in momentum

So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
3 years ago
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