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Stolb23 [73]
2 years ago
14

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6

.34). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.
(a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?
Physics
1 answer:
Rina8888 [55]2 years ago
7 0

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force \frac{mv^{2} }{r}. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

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A ball is thrown straight up with an initial velocity of 6.4 m/s. It travels for 0.64 seconds, and has a change of position of 2
Naya [18.7K]

Answer:

V = 0.9 m/s

Explanation:

The parameters given are:

Initial velocity U = 6.4 m/s

Time t = 0.64s

Height h = 2.05 m

To find the final velocity, let us use third equation of motion

V^2 = U^2 - 2gH

Since the ball is going upward, g will be negative

Substitute all the parameters into the formula

V^2 = 6.4^2 - 2 × 9.8 × 2.05

V^2 = 40.96 - 40.18

V^2 = 0.78

V = sqrt( 0.78)

V = 0.883 m/ s

V = 0.9 m/ s approximately

6 0
3 years ago
A bird carrying a fish (5kg) drops it from 107 meters in the air how fast does the fish hit the ground
notka56 [123]

Answer:

The velocity of the fish hitting the ground is , v = 45.795 m/s        

Explanation:

Given data,

The mass of the fish, m = 5 kg

The height of the bird from the surface, h = 107 m

Using the III equation of motion,

                          v² = u² + 2gs

                          <em> v = √(u² + 2gs)</em>

Substituting the values,

                           v = √(0² + 2 x 9.8 x 107)  

                              = 45.795 m/s

Hence, the velocity of the fish hitting the ground is, v = 45.795 m/s        

4 0
3 years ago
Rocks A and B are located at the same height on top of a hill. The mass of rock A is twice the mass of rock B. How does the pote
Ronch [10]

Potential energy is defined by formula

U = mgh

here

m = mass

g = acceleration due to gravity

h = height

Now here two different stones are located at same height

while mass of stone A is twice that of stone B

so here we can say potential energy of A is

U_a = (2m)gh

Similarly potential energy of B is

U_b = mgh

now if we take the ratio of two energy

\frac{U_a}{U_b} = 2

so we can say potential energy of stone A is two times the potential energy of B

7 0
3 years ago
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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When soccer players run they are using friction to propell themselves
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