A 10 kg ball is held above a building with a height of 30 m. What is the

The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity. an ostrich is run

Orlov [11]

Velocity of an object is its rate of change of the object's position per interval of time. Velocity is a vector quantity which means that it consists of a magnitude and a direction. Magnitude is represented by the speed and the direction is represented by the angle. To determine the velocity components, we use trigonometric functions to determine the angle of the components. For the north component we, use the sine function while, for the west component, we use the cosine function. We calculate as follows:
north velocity component = (16.8 m/s) (sin 54°) = 16.4 m/s
west velocity component = (16.8 m/s) (cos 54°) = 3.49 m/s

6 0

4 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

What happens when the mass of an object is greter

Paul [167]

The acceleration goes up.

4 0

3 years ago

In this example we will analyze the forces acting on your body as you move in an elevator. Specifically, we will consider the ca

dexar [7]

Answer:

The reading of the scale during the acceleration is 446.94 N

Explanation:

Given;

the reading of the scale when the elevator is at rest = your weight, w = 600 N

downward acceleration the elevator, a = 2.5 m/s²

The reading of the scale can be found by applying Newton's second law of motion;

the reading of the scale = net force acting on your body

R = mg + m(-a)

The negative sign indicates downward acceleration

R = m(g - a)

where;

R is the reading of the scale which is your apparent weight

m is the mass of your body

g is acceleration due to gravity, = 9.8 m/s²

m = w/g

m = 600 / 9.8

m = 61.225 kg

The reading of the scale is now calculated as;

R = m(g-a)

R = 61.225(9.8 - 2.5)

R = 446.94 N

Therefore, the reading of the scale during the acceleration is 446.94 N

4 0

3 years ago

A 2 kg book is pushed from rest to a final velocity of 3 m/s. The book travels 2 m. How much force was the book pushed with

Studentka2010 [4]

Explanation:

first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s

now using the 3rd law of motion

v^2=u^2+2as

0=9+2a2

a= -9/4m/s^2

now force=mass×accelration

=2kg×(-9/4)m/s^2

=4.5 N

4.5 newton force applied on the book!

✌️:)

6 0

3 years ago

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