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marin [14]
2 years ago
5

Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force

of 160 newtons on the nail. The mechanical advantage of the hammer is .
Physics
2 answers:
Dahasolnce [82]2 years ago
5 0

Answer:

Explanation:

We know that the mechanical advantage of the hammer is is given by the ratio between the force applied on the hammer and the force required to overcome the given force.Force applied = 40 NForce required= 160 NMechanical advantage =Mechanical advantage =

Hence, the mechanical advantage of the hammer is 4.

Neporo4naja [7]2 years ago
5 0

Force B / Force A

So, 160 / 40 = 4

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Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when
Arturiano [62]

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s

So, its velocity at the bottom of the ramp is 2.21 m/s.

4 0
2 years ago
A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
Roman55 [17]

Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

which is approximately 41º west of north.

6 0
2 years ago
A man stands on his balcony, 130 feet above the ground. He looks at the ground, with his sight line forming an angle of 70° with
jenyasd209 [6]

Answer:

d =  380 feet

Explanation:

Height of man = perpendicular= 130 feet

Angle of depression = ∅ = 70 °

distance to bus stop from man = hypotenuse = d = 130 sec∅

As sec ∅ = 1 / cos∅

so d = 130 sec∅    or d = 130 / cos∅

d = 130 / cos(70°)

d =  380 feet

8 0
2 years ago
If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0
2 years ago
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