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VashaNatasha [74]

4 years ago

6

What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?

1 answer:

Wewaii [24]4 years ago

3 0

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

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When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0

3 years ago

Give Example of mental flexibility

Inessa [10]

Answer:

If you ring the doorbell and no one opens the door, you'll infer that no one is home rather than continuing to ring the doorbell to an empty house. Being able to understand this and look for another solution is another example of mental flexibility.

Explanation:

6 0

3 years ago

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

DerKrebs [107]

The ball's horizontal and vertical velocities at time $t$ are

$v_x=v_{xi}$

$v_y=v_{yi}-gt$

but the ball is thrown horizontally, so $v_{yi}=0$ . Its horizontal and vertical positions at time $t$ are

$x=v_{xi}t$

$y=20.0\,\mathrm m-\dfrac g2t^2$

The ball travels 22 m horizontally from where it was thrown, so

$22\,\mathrm m=v_{xi}t$

from which we find the time it takes for the ball to land on the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it lands, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

$\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}$

7 0

3 years ago

We have all complained that there aren't enough hours in a day. In an attempt to fix that, suppose all the people in the world l

Natali5045456 [20]

Answer:

the duration of a day increases 4.96x10^-11 s

Explanation:

According the exercise:

R=radius of Earth=6.37x10^6 m

mE=mass of Earth=5.97x10^24 kg

m=average mass of people=55 kg

n=number of population=7x10^9

M=total mass=n*m=7x10^9*55=3.85x10^11 kg

v=speed=2.5 m/s

The moment of inertia of population is:

I=(2/3)*M*R^2=(2/3)*3.85x10^11*(6.37x10^6)=1.04x10^25 kg*m^2

The time taken per revolution is:

T=2πR/v=(2π*6.37x10^6)/2.5=1.6x10^7 rev/s

The angular speed is:

w=2π/T=2π/1.6x10^7=3.9x10^-7 rad/s

The angular momentum of population is equal to:

L1=I*w=1.04x10^25*3.9x10^-7=4.08x10^18 kg*m^2/s

The angular momentum of Earth is equal to:

L2=I*w=((2/5)*me*R^2)*(2π/24)=((2/5)*5.97x10^24*(6.37x10^6)^2)*(2π/(24*60*60))=7.1x10^33 kg*m^2/s

The change in length of the day is equal to:

T´=T*(L1/L2)=(24*60*60)*(4.08x10^18/7.1x10^33)=4.96x10^-11 s

8 0

3 years ago

What is the elevation of Y and Z? Y=3250, Z=2950 Y=3200, Z=2900 Y=3250, Z=2900 Y=3000, Z=2850

TiliK225 [7]

I think the elevation of Y and Z are the following:
<span>Y=3200,
Z=2900 </span>

8 0

3 years ago