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ziro4ka [17]
3 years ago
13

A pendulum is 0.760 m long, and the bob has a mass of 1,00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. The tens

ion is greater than the weight of the bob because Multiple Choice 0 its horizontal component is increasing from zero. O the bob has a downward acceleration, so the net Fy must be downward O O the bob has zero acceleration. O ( the bob has an upward acceleration, so the net Fy must be upward Required information A pendulum is 0.760 m long, and the bob has a mass of 1.00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. What is the tension in the string at the bottom of the swing?
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

  T = 13.17 N

the correct one is there is a digested centripetal acceleration towards upward

Explanation:

For this exercise we can use Newton's second law at the bottom of the pendulum's path

we will assume that the upward direction is positive

              T - W = m a

in this case it is describing a circle the acceleration is central

              a = v² / R

where the radius of the trajectories equals the length of the pendulum

             R = L

we substitute

            T = mg + m v² / L

             T = m (g + v² / L)

When reviewing the different statements, the correct one is there is a digested centripetal acceleration towards upward

let's calculate the tension of the rope

             T = 1 (9.8 + 1.6 2 / 0.760)

             T = 13.17 N

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Amiraneli [1.4K]

Answer:

The push switch is closed and the door unlocks. ... The switch completes the circuit and a current flows through the coil. The coil becomes an electromagnet and its iron core becomes magnetised. The iron bolt is attracted to the electromagnet.

Explanation:

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5 0
3 years ago
A What quantity is equal to​
natulia [17]

Answer:

Option B is correct

Explanation:

acceleration is equal to rate of change of velocity.

6 0
3 years ago
A Light spiral spring is loaded with
kakasveta [241]

Answer:

0.63 s

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 50 g

Extention (e) = 10 cm

Period (T) =?

Next, we obtained 50 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

50 g = 50 g × 1 Kg / 1000 g

50 g = 0.05 kg

Next, we shall convert 10 cm to m. This is illustrated below:

100 cm = 1 m

Therefore,

10 cm = 10 cm × 1 m / 100 cm

10 cm = 0.1 m

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass = 0.05 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 0.05 × 9.8

F = 0.49 N

Next, we shall determine the spring constant of the spring.

Extention (e) = 0.1 m

Force (F) = 0.49 N

Spring constant (K) =?

F = Ke

0.49 = K × 0.1

Divide both side by 0.1

K = 0.49 /0.1

K = 4.9 N/m

Finally, we shall determine the period as follow:

Mass = 0.05 Kg

Spring constant (K) = 4.9 N/m

Pi (π) = 3.14

Period (T) =?

T = 2π√(m/k)

T = 2 × 3.14 × √(0.05 / 4.9)

T = 6.28 × √(0.05 / 4.9)

T = 0.63 s

Thus, the period of oscillation is 0.63 s

4 0
3 years ago
You place a 10 kg block on a ramp with an angle of 20 degrees. You push the block up the ramp giving it an initial velocity of 1
Ainat [17]

Answer:

L = 15.97 m

Explanation:

Given:-

- The mass of the block, m = 10 kg

- The inclination of ramp, θ = 20°

- The initial speed, Vi = 15 m/s

- The coefficient of friction u = 0.4

Find:-

find the total distance the block travels before it turns around and slides back down the ramp.

Solution:-

- The total distance travelled by the block up the ramp is defined when all the kinetic energy is converted into potential energy and work is done against the friction. Final velocity V2 = 0.

- Develop a free body diagram of the block. Resolve the weight "W" of the block normal to the surface of ramp. Then apply equilibrium condition for the block in the direction normal to the surface:

                                N - W*cos( θ ) = 0

Where, N : The contact force between block and ramp.

                                N = m*g*cos ( θ )

- The friction force (Ff) is defined as:

                               Ff = u*N

                               Ff = u*m*g*cos ( θ )

- Apply the work-energy principle for the block which travels a distance of "L" up the ramp:

                               K.E i = P.E f + Work done against friction

Where,  K.E i = 0.5*m*Vi^2

             P.E f = m*g*L*sin( θ )

             Work done = Ff*L

- Evaluate "L":

                        0.5*m*Vi^2 = m*g*L*sin( θ ) + u*m*g*cos ( θ )*L

                        0.5*Vi^2 = g*L*sin( θ ) + u*g*cos ( θ )*L

                        0.5*Vi^2 = L [ g*sin( θ ) + u*g*cos ( θ ) ]

                        L = 0.5*Vi^2 / [ g*sin( θ ) + u*g*cos ( θ ) ]

                        L = 0.5*15^2 / [ 9.81*sin( 20 ) + 0.4*9.81*cos ( 20 ) ]

                        L = 15.97 m

7 0
3 years ago
The turbines can be seen inside this hydroelectric dam. Why are they located at that particular height?
Yakvenalex [24]

Answer:

3

Explanation:

the answer is number three

5 0
4 years ago
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