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UkoKoshka [18]
3 years ago
5

A brick is dropped from a big scaffold. What is its velocity after 4.0s?

Physics
1 answer:
Flura [38]3 years ago
3 0

The velocity of the brick is 39.2 m/s downward

Explanation:

The motion of the brick is a free fall motion, since the object is affected only by the force of gravity. Therefore, it has a uniformly accelerated motion towards the ground, with constant acceleration of g=9.8 m/s^2.

So, we can find its velocity using the suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the brick in this problem (taking downward as positive direction)

u = 0 (it is dropped from rest)

a=g=9.8 m/s^2

Therefore, its velocity after t = 4.0 s is:

v=0+(9.8)(4.0)=39.2 m/s

Downward, because the sign is positive.

Learn more about free fall motion:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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3 0
2 years ago
How many mm are in a micro-m?
Dahasolnce [82]

For this case we have that by definition, a micrometer is equivalent to a thousandth of a millimeter, that is, 0.001 millimeters.

Then, in other words, we have 0.001 millimeters in a micrometer.

Answer:

In a micrometer there are 0.001 millimeters.

4 0
3 years ago
A car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The diameter of a tire is 80.2 cm. Find the numb
BaLLatris [955]

By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

  • Initial velocity U = 0
  • Final velocity V = 24.3 m/s
  • Time t = 9.1 s

If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

a = 24.3/9.1

a = 2.67 m/s^{2}

Third Equation of motion

V^{2} = U^{2} + 2aS

Substitute all the necessary parameters

24.3^{2} = 0 + 2 x 2.67 x S

590.49 = 5.34S

S = 590.49 / 5.34

S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

convert it to meter

r = 40.1/100 = 0.401 m

The Circumference of the tire = 2\pir

Circumference = 2 x 3.143 x 0.401

Circumference = 2.52 m

Assuming no slipping, number of revolutions = 110.58/2.52

Number of revolutions = 43.89 rev.

Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

4 0
2 years ago
A marble rolls along the floor with a constant velocity of 6 m/s. How far will it have gone after 130 seconds
SCORPION-xisa [38]

Answer:

0.046

Explanation:

displacement = velocity/ time

d = 6m/s / 130s

d = 0.046m

5 0
3 years ago
If planet A is four times as far from the sun as planet C, then the period of its orbit will be__ times as long
Burka [1]
Within the system of the same star, the period of a planet's orbit is
proportional to the 3/2 power of its distance from the central body.
(Kepler's empirical third law of planetary motion, promoted to being
etched in stone by Newton's gravitation.)

(4) ^ 3/2 = <em>8 times</em> as long.
4 0
4 years ago
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