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2 years ago

I need help with 03.01 Playing With Others – President's Challenge Goals

Physics

1 answer:

galben [10]2 years ago

4 0

Answer:

is this like a book orsomthing

Explanation:

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As the elephant falls from 10 m does it lose or gain KE? Explain.

ivolga24 [154]

He loss KE hope this helps

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3 years ago

Read 2 more answers

You own a high speed digital camera that can take a picture every 0.5 seconds. You decide to take a picture every 0.5 seconds of

Vesnalui [34]

-- There is no need to develop the pictures. They are available immediately in a digital camera.

-- There is no change in the teacher from one picture to the next.

-- The distance the watermelon falls from the teacher in each new picture is more in each picture than in the picture before it. (C)

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2 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

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3 years ago

Any Answer Would Be Appreciated

-Dominant- [34]

Answer:

last one.Condution and convection are equzlly efficentmethods of heat transfer.

Explanation:

3 0

2 years ago

a stone is projected vertically up from the top of a tower 73.5m with velocity 24.5 m/s . find the time taken by the stone to re

hoa [83]

The stone's altitude at time $t$ is given by

$y=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\dfrac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity. The stone reaches the ground when $y=0$ :

$0=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=7.11\,\rm s$

6 0

2 years ago