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3 years ago

What is the net force of a 25 g object with an acceleration of 3 m/s^2? 100PTS

Physics

2 answers:

gayaneshka [121]3 years ago

5 0

Explanation:

Force=Mass×acceleration

force=25×3

force=75N

sertanlavr [38]3 years ago

5 0

Answer:

0.75 N

Explanation:

F=ma

mass change to kg -> 25 g to 0.25 Kg

0.25x3 =0.75 N

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In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra

shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0

4 years ago

Which of the following statements is TRUE about updating the exposure control plan?

iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

Updates must have the reflection of changes in tasks as well in procedures.

Updates must reflect changes in positions that affect occupational exposure.

Updates must have the cost of PPE that is needed and necessary to reduce exposure

An exposure control plan can be regarded as the framework for compliance between the employer and the workers.

This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

This plan gives hope to workers in term of protection when working with their Employer.

There are some elements that is associated with Exposure Control Plan, and theses are;

Health hazards as well as risk that is attributed to each product in the worksite.
Statement of purpose.
procedures and practices in a written form
Responsibilities from the Manager, CEO, designated resources and employer.

Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.

brainly.com/question/1203927?referrer=searchResults

3 0

2 years ago

The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a

Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

with m the mass , a the acceleration and $\sum\overrightarrow{F}$ the net force forces that is:

$(F-f)$ (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

$(F-f)=ma$

solving for a:

$a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}$

Now let's use the Galileo’s kinematic equation

$Vf^{2}=Vo^{2}+2a\varDelta x$ (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and $\varDelta x$ the time took to achieve that velocity, solving (3) for $\varDelta x$ :