If a pitcher throws a baseball 10 m in 30 seconds - what would be the average speed of the baseball

For this situation to demonstrate a balanced force, how much force must Omar apply? (In the picture, Sam is pushing a box to the

11111nata11111 [884]

The answers 10N as it equals it out

3 0

3 years ago

Read 2 more answers

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

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

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0

3 years ago

A 72-tooth gear is connected to a 12-tooth gear. If the large gear makes one complete turn, how many turns

viktelen [127]

You divide 72 by 12 to get 6.

8 0

3 years ago

The displacement of a wave traveling in the positive x-direction is y(x, t)|= (3.5 cm)cos(2.7x − 92t), where x is in m and t is

zubka84 [21]

Answer

Given,

y(x, t) = (3.5 cm) cos(2.7 x − 92 t)

comparing the given equation with general equation

y(x,t) = A cos(k x - ω t)

A = 3.5 cm , k = 2.7 rad/m , ω = 92 rad/s

we know,

a) ω =2πf

f = 92/ 2π

f = 14.64 Hz

b) Wavelength of the wave

we now, k = 2π/λ

2π/λ = 2.7

λ = 2 π/2.7

λ = 2.33 m

c) Speed of wave

v = ν λ

v = 14.64 x 2.33

v = 34.11 m/s

5 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago