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Svetllana [295]
3 years ago
9

Which of the following is true about the following lever?

Physics
2 answers:
NISA [10]3 years ago
7 0

Answer:

Explanation:

i think the answer is c... don't quote me on that...  i'm not 100% sure...

Natali [406]3 years ago
4 0
Oh jk haha LOL bye hahaha
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A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to
nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to  the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

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How do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate?
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What is a loose icy body with a long narrow orbit?
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You can only score 1 point when shooting a free throw. Group of answer choices True False
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Answer:

2 points

Explanation:

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Read 2 more answers
A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable
Nuetrik [128]

Answer:

a) 29062.125 N·m

b) 0 N·m

c) Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

8 0
3 years ago
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