When a falling object reaches terminal velocity, the net force acting on it is

What planet are people working on so they can move there

Nata [24]

Answer:

mars

Explanation:

just search it up

8 0

3 years ago

Read 2 more answers

A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

3 years ago

If you have a gold brick that is 2.0 cm by 3.0 cm by 4.0 cm and has a density of 19.3 g/ml, what is its mass

34kurt

Answer:

It is

Explanation:

1 Answer. The volume is 37.0 cm3Au .

8 0

3 years ago

PLEASE HELPPPPP

Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (zero movement) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

graph B, and

graph D.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0

2 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$