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faltersainse [42]
3 years ago
5

The unit used to measure electric current is the ampere (AA). Now, assume that the current delivered at a wall socket reaches th

e value 3.8 AA ten times in a time interval of 0.17 ss. What is the period TTT with which the current at the wall socket changes?
Physics
1 answer:
Lostsunrise [7]3 years ago
8 0

The period T with which the current at the wall socket changes is 0.017 s.

It also ranges from 0.155 s to 0.0189 s since the cycles is not given in the question.

<u>Explanation</u>:

The current reaches 3.8 A once for every cycle. So there must have been 10 cycles (10 periods) in 0.17 s.  

Using 'T' for period:

                                10 T = 0.17  

                                      T = 0.17 / 10  

                                     T  = 0.017 s.

The question doesn't tell about the cycles and so we assume it as 10 cycles. If there are 9 cycles due to the current 3.8 A at the exact beginning and end of 0.17 s. This would make the period T as

                              0.17 / 9 = 0.0189 s approx.

Likewise, if the 3.8 A occurs at the end of the 1st cycle, 11 cycles would have passed during 0.17 s.

In that case the period = 0.17 / 11 = 0.155 s approx.

So T could be in the range from 0.155 s to 0.0189 s.

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A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of frictio
Leya [2.2K]

Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

3 0
3 years ago
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8. A crate of bananas weighing 3000 N is shipped from South America to New York, where it
LiRa [457]

Answer:

Mechanical advantage = 15

Explanation:

Given the following data;

Output force = 3000N

Input force = 200N

To find the mechanical advantage;

Mechanical advantage = output force/input force

Substituting into the equation, we have

Mechanical advantage = 3000/200

Mechanical advantage = 15

3 0
2 years ago
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0
2 years ago
The activation of clotting factors will be blocked.
dedylja [7]

With the blocking of activation of clotting factors, the rate of conversion of fibrinogen to fibrin will decrease to a huge extent and this will prevent the clot formation.

Option A

<h3><u>Explanation:</u></h3>

The process of stopping of flow of blood through any wound by formation of a clot is known as blood clotting. The clot in blood is formed by conversion of the fibrinogen protein into its polymer form fibrin which forms a meshwork.

The conversion of fibrinogen to fibrin requires a lot of enzymes and factors present which is required one by one, known as the Cascade theory. Total of 13 factors are required, where there are prothrombin, thromboplastin, and different other factors. Inactivation of any of the 13 factors will lead to less conversion of fibrinogen to fibrin, thereby the rate of conversion will highly decrease.

7 0
2 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
3 years ago
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