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EastWind [94]
4 years ago
9

Questions for 1.21 physics lab report

Physics
1 answer:
Lapatulllka [165]4 years ago
6 0
Ok cool dude bro I just need to answer a question
You might be interested in
A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin
Sidana [21]

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

8 0
3 years ago
A dad takes his kids to their school just 8.0 miles down the road but with traffic it takes him 30 minutes and the fastest he ca
VARVARA [1.3K]

Answer:C 24 mi/hr

Explanation:

8 0
3 years ago
Why do you think matter so varied
Natali [406]

Answer:

Matter is anything that has mass and takes up space. A mixture is matter that can vary in composition.....because, the substances in a heterogeneous mixture are not evenly mixed, two samples of the same mixture can have different amounts of the substance....

Hope this answer help u!!!!

Pls mark as brainlist!!!

6 0
4 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
3 years ago
On Ella's fifth birthday, her parents told her she had to stop whining or no one would want to play with her. She tried very har
Kisachek [45]
I think the answer to this problem I believe would probably be B. learned optimism. I think it's the closest answer...I THINK
3 0
3 years ago
Read 2 more answers
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