Simplify log√8÷log√8

A baseball (m=140g) traveling 32m/s moves a fielders

posledela

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average force exerted by the ball on the glove. It is solved using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

F = mg

$\dfrac{1}{2}mv^2=Fh$

$F=\dfrac{mv^2}{2h}$

$F=\dfrac{0.14\times (32)^2}{2\times 0.25}$

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

7 0

3 years ago

What is the wavelength of the waves you create in a swimming pool if you splash your hand at a rate of 2.00 Hz and the waves pro

Valentin [98]

Answer:

The wavelength of the waves created in the swimming pool is 0.4 m

Explanation:

Given;

frequency of the wave, f = 2 Hz

velocity of the wave, v = 0.8 m/s

The wavelength of the wave is given by;

λ = v / f

where;

λ is the wavelength

f is the frequency

v is the wavelength

λ = 0.8 / 2

λ = 0.4 m

Therefore, the wavelength of the waves created in the swimming pool is 0.4 m

3 0

3 years ago

Obtain a volume of 12.5 millimeters of liquid in the diagram

tiny-mole [99]

Answer:

D remove 1.5 ML of liquid.

Explanation:

5 0

2 years ago

Read 2 more answers

A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

Anna35 [415]

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

$h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2$

The direction of the acceleration of the ball is downwards

Learn more here: brainly.com/question/15407740

4 0

2 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago

Simplify log√8÷log√8​

Simplify log√8÷log√8