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3 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

Physics

1 answer:

lawyer [7]3 years ago

4 0

I really think it could be a 23 kg

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Atoms with the same number of protons but with different electrical charges _____.

mash [69]

Answer:

Atoms with the same number of protons but with different electrical charges are different ions

Explanation:

Ions are defined as those atoms or molecules which carry charge

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3 years ago

Read 2 more answers

I WILL GIVE YOU BRAINLIEST IF YOU ANSWER THOS QUESTION IN THE NEXT 5 MINUTES!!

schepotkina [342]

Answer:

the blue shopping cart.

Explanation:

The blue shopping cart doesnt have to worry about running someone over in the front. The red one does, so it slows down more.

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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.

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

3 0

3 years ago

According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s

Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

$\tau_{net} = 0$

now we have

$\frac{dL}{dt}= 0$

now we also know that

$Area = \frac{1}{2}r^2d\theta$

so rate of change in area is given as

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}$

so we will have

$\frac{dA}{dt} = \frac{1}{2}r^2\omega$

$\frac{dA}{dt} = \frac{L}{2m}$

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0

3 years ago

The plain flight covers 400 miles in the first hour. As you approach the landing site, it takes you an hour to cover the last 20

tatyana61 [14]

that is an example of negative acceleration because it is slowing down

3 0

3 years ago