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3 years ago

How long does it take a person to skate the width of a hockey rink (85 feet) at a constant speed of 15 feet per second?

1 answer:

3 0

S(travel distance)=85 ft
v (velocity)=15 ft/s
-----------------------------------
t (time)=?

Calculate the time with the formula for the velocity:
v=S/t
t=S/v
t=85 ft/(15 ft/s)
t=5.666s

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A 1380 kg car starts from rest at the

Ulleksa [173]

Answer: 72200

Explanation:

First you must find the height for this is on an inclined hill using:

h=Lsin(angle) —> 28.0sin(11.0) = 5.34

Now you would just use the PE equation (mgh) because you are finding ME and when you starting from the top KE=0, showing that what ever answer you get from PE would equal the same for ME.

Using mgh:

m=1380

g=9.80

h=5.34

(1380)(9.8)(5.34)

=72218.16

*Rounding to the 3rd=72200

Hope this helps :)

6 0

2 years ago

2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so

evablogger [386]

The answer is (a) hope it helped!<3

5 0

2 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0

3 years ago

How much energy is stored in the electric field of a 50-μm-diameter cell with a 7.0-nm-thick cell wall whose dielectric constant

Nady [450]

:<span> </span><span>Under the assumption that a cell is made up of two concentric spheres you find the surface are of the inside sphere which will be your A.

You already have your separation and dielectric constant so just use the formula you stated towards the end of your question and you get 8.93x10^-11 Farads which is about 89pF</span>

7 0

2 years ago

Read 2 more answers

A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu

Katyanochek1 [597]

Answer:

a) V_f = 25.514 m/s

b) Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

V_f = V_i + a*t

V_f = 20.5 j + 0.31 i *49

V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

V_f = sqrt ( 20.5^2 + 15.19^2)

V_f = sqrt(650.9861)

V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

tan(Q) = (V_y / V_x)

Q = arctan (20.5 / 15.19)

Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0

3 years ago