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Svetllana [295]
3 years ago
6

A car accelerates uniformly in a straight line with acceleration 10m/s 2 and travels 150m in a time interval of 5s. How far will

it travel in the next 5s?​
Physics
1 answer:
spayn [35]3 years ago
6 0

Explanation:

Given:

Acceleration of car = 10 m/s

Distance travelled in 5 sec = 150 m

To find:

Distance travelled in the next 5 seconds

Concept:

There are 2 ways to app this kind of questions .

Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.

Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.

Calculation:

v² = u² + 2as

=> ( u + at)² = u² + 2as

=> u² + 2uat + a²t² = u² + 2as

=> 2uat + (at)² = 2as

=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)

=> 100u = 3000 - 2500

=> 100u = 500

=> u = 5 m/s

Distance travelled in 10 seconds :

s = ut + ½at²

=> s = (5 × 10) + ½(10)(10)²

=> s = 50 + 500

=> s = 550 m

Distance travelled in the 2nd half will be :

d = 550 - 150

=> d = 400 m

So final answer is :

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W and X

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8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included  with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, T_P = 30 N

The tension in rope Q, T_Q = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, \Sigma F_y = T_P_y + T_Q_y + W = 0

∴ W = -(T_P_y + T_Q_y)

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, \Sigma F_x = T_P_x + T_Q_x  = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

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3 0
3 years ago
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
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Answer:

Fnet = F√2

Fnet = kq²/r² √2

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A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

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5 0
3 years ago
Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the
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Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

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t_B=0.45s

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(b)d_B=v \times t_B

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v= speed of sound

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(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m

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