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Mekhanik [1.2K]
3 years ago
10

A person is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follo

ws: P is the upward force the person exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the person is trying unsuccessfully to lift the crate?
Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

P + C = W if in equilibrium?

P + C - W = 0; P + C = W

Explanation:

Action and reaction are equal and opposite according to newton's third law of motion. a force is that which tends to change a body's state of rest or uniform motion in a straight line, recall

P is the upward force the person exerts on the crate,

C is the vertical contact force exerted on the crate by the floor,

W is the weight of the crate

let the sum of upward forces be on the left hand side of the equation and the sum of downward forces on the right hand side.

P+C=W

P+C-W=0 if they are in equilibrium

the forces are related by the above.

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Answer:

Pressure on the molten rock lessens and the gases dissolved in rock can bubble and expand rapidly causing violent eruptions.

Explanation:

that's just how it works lol. hope this helps :]

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3 years ago
A railroad car with a mass of 30,000 kg is moving at 2.0 m/s when it runs into an at-rest freight car with an equal mass. The ca
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The total momentum of the system is preserved through the collision.

Note that momentum is
P = m*v
where m = mass
v = velocity.

Initial momentum:
P1 = (30000 kg)*(2 m/s) = 60000 (kg-m)/s for the moving car
P2 = 0 for the starionary car.

Final momentum:
P3 = (30000 + 30000)*v = 60000v (kg-m)/s

Because momentum is preserved,
P3 = P1 + P2
60000v = 60000
v = 1 m/s
The final velocity is 1 m/s.

Answer: 1.0 m/s
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3 years ago
9. Suppose you ride your Segway to the library traveling at 0.5 km/min. It takes you 25
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Answer:

12.5 km

Explanation:

0.5 x 25 = 12.5

8 0
3 years ago
Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h
madreJ [45]

Answer:

(a) a = 1.875 m/s²

(b)T = 1.575 N

(c)T increase

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

Forces acting on the block  A

WA: Weight of of the A block : In vertical direction  downaward (-y)

NA : Normal force of the A block :In vertical direction  upaward (+y)

F= 5.7 N  In in the direction parallel to the movement of the blocks (+x)

T :  tension of the string: In in the direction (-x)

Forces acting on the block  B

WB: Weight of the B block: In vertical direction  downaward (-y)

NB : Normal force of the B block :In vertical direction  upaward (+y)

T :  tension of the string: In in the direction (+x)

Data

mA = 2.2 kg

mB = 0.84 kg

(a) Magnitude of the acceleration  of the blocks

Newton's second law to B block:

∑Fx = m*a

T = (0.84)*a Equation (1)

Newton's second law to A block:

∑Fx = m*a

5.7 - T = (2.2)*a  We replace T of the Equation (1)

5.7 - (0.84)*a = (2.2)*a  

5.7 = (2.2)*a + (0.84)*a  Equation (2)

5.7 =(3.04)*a

a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the  Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T  increase

4 0
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A fisherman sees his boat moving up and down, owing to waves on the water. It takes 2.5 s for the boat to travel from its highes
nata0808 [166]

Answer:

The waves are traveling in a speed of

v=1.20 m/s

Explanation:

The fisherman see a wave at 6m apart so that's the wavelength

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Now the period is the time between two successive waves so

t=2*2.5s=5s

The velocity of the waves is describe by:

v=f*L

f=\frac{1}{t}=\frac{1}{5s}=20s^{-1}

v=0.20s^{-1}*6m

v=1.20 \frac{m}{s}

7 0
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