A bowling ball is pushed with a force of 22.0N and accerlates at 5.5 m/s square. What is the mass of the bowling ball

Tomas wants to paint his bedroom. He goes to the store and buys two cans of paint.

GuDViN [60]

C an inclined plate
youre welcome

8 0

2 years ago

On Ella's fifth birthday, her parents told her she had to stop whining or no one would want to play with her. She tried very har

Kisachek [45]

I think the answer to this problem I believe would probably be B. learned optimism. I think it's the closest answer...I THINK

3 0

2 years ago

Read 2 more answers

If the wavelength of a sound wave increases and the frequency of the sound wave does not change what happens to the speed of the

salantis [7]

If the wavelength of a sound wave increases and the frequency of the sound wave does not change, the speed of the wave will increase.

Ans: D

Explanation

The sound wave speed is given by E=fλ, where f indicates its frequency and λ indicates its wavelength.

From the equation, it is evident that the sound speed is proportional to both frequency and wavelength.

Here, as wavelength increases, wave speed increases provided there is no change in frequency.

3 0

2 years ago

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

a

The height is $H = 6.74 \ m$

b

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$

So

$-1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2$

=> $-4.9t^2 + 9.6418t + 1 = 0$

using quadratic formula to solve the equation we have

$t = 2.07 \ s$

Generally distance covered along the horizontal is

$D = v cos (40) * 2.07$

=> $D = 15 cos (40) * 2.07$

=> $D = 23.74 \ m$

7 0

2 years ago

List two applications of electrostatics.

Tems11 [23]

Atmospheric electricity and storms,electric current in a vacuum,spark discharge,electrostatic control filters and industrial electrostatic separation <- those are just a few

4 0

2 years ago