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Dafna1 [17]
3 years ago
7

a 2-kg object is dropped from a height of 1000 m. What is the force of air resistance on the object when it reaches terminal vel

ocity
Physics
1 answer:
strojnjashka [21]3 years ago
5 0
It stops accelerating when the air resistance is equal to its weight.
That's (m•g)

= (2 kg) • (9.8 m/s^2)

= 19.6 newtons
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VMariaS [17]

Answer:

Zeros that follow non-zero numbers and are also to the right of a decimal point are significant.

Explanation:

For example:

0.300 has 3 significant figures.

5.400 has 4 significant figures.

4 0
2 years ago
Which of the following is a possible use for nuclear energy?
belka [17]
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5 0
4 years ago
Read 2 more answers
A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4
LekaFEV [45]

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

4 0
3 years ago
A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the bl
KengaRu [80]

Answer:

a) \ d_2=d_1\\b) \ d_2=4d_1

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

6 0
4 years ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
4 years ago
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