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Pepsi [2]
2 years ago
14

A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction

al force of 560 N. Find the mass of the worker.

Physics
1 answer:
aleksley [76]2 years ago
5 0

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

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Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

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1 year ago
Which trailer has more downward pressure where it attaches to the car?
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The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

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2 years ago
Why do most objects tend to contain nearly equal numbers of positive and negative charges?
mart [117]
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Imagine you derive the following expression by analyzing the physics of a particular system: a=gsinθ−μkgcosθ, where g=9.80meter/
BARSIC [14]

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

8 0
2 years ago
If a metallic wire of cross sectional area 3.0 ´ 10-6 m2 carries a current of 6.0 A and has a mobile charge density of 4.24 ´ 10
elena-14-01-66 [18.8K]

Answer:

The drift velocity is v_d=2.9\times 10^{-4}\ m/s.

Explanation:

Given :

Area of metallic wire, A = 3\times 10^{-6}\ m^2.

Current through wire , I=6 \ A.

Mobile charge density , n=4.24\times 10^{28} \ carriers/m^3.

Charge value , e=1.6\times 10^{-19}\ C.

We need to find drift velocity , v_d.

Now, we know :

I=neAv_d

Therefore, v_d=\dfrac{I}{neA}

Putting all given values in above equation we get,

v_d=\dfrac{6}{4.24\times 10^{28}\times 1.6\times 10^{-19} \times 3 \times 10^{-6}}

v_d=2.9\times 10^{-4}\ m/s.

Hence, this is the required solution.

8 0
2 years ago
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