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3 years ago

11

Calculate the change in velocity of a 0.070kg tennis ball hit by Serena with a force of 140 N over 0.020 s

1 answer:

Brilliant_brown [7]3 years ago

6 0

V=at and a=F/m

140/.070 = 2000m/s^2

2000*.020 = 40m/s

The ball’s velocity increased by 40m/s.

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Two waves, each with intensity 40 db, interfere constructively. what is the intensity of the combined waves, in db?

Nikitich [7]

I'm not really sure it's been awhile since I did this.

4 0

3 years ago

A sprinter speeds up to 3 m/s during the last 2 seconds of the race with an

quester [9]

Answer:

<em>The initial speed of the sprinter was 2.2 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

The following relation applies:

$v_f=v_o+at$

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The sprinter speeds up from an unknown initial speed to vf=3 m/s in t=2 seconds with an acceleration of $a=0.4~m/s^2$ .

To find the initial speed, we solve the equation for vo:

$v_o=v_f-at$

Substituting the values:

$v_o=3-0.4*2$

$v_o=3-0.8$

$v_o=2.2~m/s$

The initial speed of the sprinter was 2.2 m/s

4 0

3 years ago

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

6 0

3 years ago

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago

If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth

Ymorist [56]

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

F_b = Y * V_timber

F_b = 9810*0.15*0.15*x

F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

(M)_p = 0

W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

670*0.5*3.6 - x^2 * 0.5*226.7 = 0

x^2 = 1206 / 113.35

x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

sin(Q) = y / x

sin(Q) = 2.1 / 3.26

Q = sin^-1 (0.6441718)

Q = 40.1 degrees

5 0

3 years ago