Benjamin, a transport technician, was punched by a patient in the

What causes charges to move in a circuit?

ipn [44]

A )voltage
(i.e differance in electricity field)

7 0

3 years ago

A tennis player swings her 1000 g racket with a speed of 11 m/s. She hits a 60 g tennis ball that was approaching her at a speed

shusha [124]

Answer:

- 3.72 Ns.

9.44 m/s

Explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

final velocity of ball, v = - 44 m/s

Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

= - 2.64 - 1.08 = - 3.72 Ns

Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 = 1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .

3 0

3 years ago

A crane moves a 250 kg scoreboard from the ground to the height of 100 m. What is the work done on the scoreboard?

Mazyrski [523]

If you’re doing potential and kinetic energy then the answer is potential.

7 0

2 years ago

Read 2 more answers

A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 7.0

krok68 [10]

Answer:

(a) T=6.07s

(b) $x_{max|4.0s}=3.62cm$

Explanation:

For Part (a)

The initial amplitude is given as A=7.0 cm

Apply the equation $x_{max}(t)=Ae^{-t/T}$ with $x_{max}(7.6s)=2.0cm$ we have:

$x_{max}(t)=Ae^{-t/T}\\2.0cm=(7.0cm)e^{-7.6s/T}\\T=-\frac{7.6s}{ln(\frac{2.0cm}{7.0cm} )} \\T=6.07s$

For Part (b)

Apply $x_{max}(t)=Ae^{-t/T}$ with t=4.0s and T=6.07s we have

$x_{max}(t)=Ae^{-t/T}\\x_{max}(4.0s)=(7.0cm)e^{-4.0/6.07}\\x_{max|4.0s}=3.62cm$

8 0

3 years ago

Use the information presented in the graph to answer

Angelina_Jolie [31]

1) Segments A and C

2) Segment C

3) 40 m/s

4) 0

Explanation:

The graph is missing: find it in attachment.

1)

Here we want to find which segments show acceleration.

Acceleration is defined as the rate of change of velocity of an object; mathematically:

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval

On a velocity-time graph, an object has acceleration if its velocity, v, changes versus time.

In this problem, we see that this situation occurs in two segments:

- In segment A, where the is a change in velocity ( $\Delta v>0$ ), so there is an acceleration

- In segment C, where the is also a change in velocity ( $\Delta v>0$ ), so there is an acceleration

B)

Here we want to find in which segment the object is slowing down.

For an object to slow down, the final velocity must be less than the initial velocity, which means that the change in velocity must be negative:

$\Delta v = v-u$

where

v is the final velocity

u is the initial velocity

For the graph in the problem, we see that:

- For segment A, the final velocity is greater than the initial velocity, so the object is speeding up

- For segment B, the final velocity is equal to the initial velocity, so the object is neither speeding up nor slowing down

- For segment C, the final velocity is less than the initial velocity, so the object is slowing down

So, the segment in which the object is slowing down is segment C.

C)

The velocity of an object is defined as the displacement covered per unit time:

$v=\frac{d}{\Delta t}$

where

d is the displacement

$\Delta t$ is the time elapsed

On a velocity-time graph, the velocity is represented on the y-axis, while the time is represented on the x-axis.

Therefore, it is possible to read directly the value of the velocity from the y-axis.

For segment B, we see that the velocity is constant, and its value is

$v_B=40 m/s$

D)

As we stated before, the acceleration of an object is the rate of change of its velocity:

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval

Here we want to find the acceleration for segment B. For this segment, we observe that:

- The velocity does not change, so $\Delta v = 0$

- The time interval is $\Delta t = 9 s - 4 s = 5 s$

Therefore, the acceleration in segment B is:

$a_B=\frac{0}{5s}=0$

6 0

3 years ago