What is the weight of a 20 kg box on the Earth?<br> A) 2 N <br> B) 20 N<br> C) 200 N

A pitcher throws a 0.143-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

o-na [289]

Answer:

200N appox

Explanation:

Step one:

Given data

mass of ball= 0.143kg

initial velocity of ball u= 42m/s

final velocity of the ball= 49m/s

time of impact= 0.005s

Step two

From the relation Ft=mΔv

we can find the average force as

F=mΔv/t

substitute

F=0.143*(49-42)/0.005

F=0.143*7/0.005

F=1.001/0.005

F=200.2

F= 200N appox

The average force is 200N

5 0

3 years ago

I am having trouble with writing a 2 paragraph summary about this vidio can someone help me

4vir4ik [10]

Answer:

where is the video?

Explanation:

8 0

2 years ago

A car is traveling to the right with a speed of 29\,\dfrac{\text m}{\text s}29

ryzh [129]

Burn us bud ijdghdhd jdnjdbd handguns jsbgd bdhdhfbdj hdjdhdhdh

4 0

3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

2 years ago

I have a special thermos that has a vacuum chamber between my hot soup and its outside wall. Because my soup is isolated from it

bagirrra123 [75]

Answer:

C. My soup will cool down because of radiation.

Explanation:

Since, there is vacuum between hot soup and its outside wall, then heat can not flow through conduction and convection.

The heat then only flows through radiation.

Therefore, the soup will not cool down because of convection or conduction by because of radiation.

7 0

3 years ago

What is the weight of a 20 kg box on the Earth? A) 2 N B) 20 N C) 200 N