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2 years ago

What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?

Physics

1 answer:

vitfil [10]2 years ago

3 0

Answer:

Option C

Explanation:

v= u + at

20 = 5 + a(5)

15= a(5)

a= 3 m/s²

Force = mass × acceleration

= 10 × 3

= 30 N

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When considering the gas laws, the kelvin temperature scale must be used. the reason for this is that the kelvin scale is direct

suter [353]

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3 years ago

Power of an electric motor is 1 h.p. what does it mean ?

const2013 [10]

Answer: HP = Horse Power.

Explanation: it is the unit given to tell the motor's particular power and 1hp = 746 watts.

8 0

2 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

Which of the following represents a fully plastic collision? 1. Two billiard balls hit each other then move in opposite directio

Nutka1998 [239]

Answer:

Option (4)

Explanation:

There are two types of collision.

Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.

Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.

Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.

3 0

3 years ago

A mass accelerates uniformly when the resultant force on it is?

Elis [28]

. . . . . not zero .

Note: "... unbalanced" would be a terrible answer.

5 0

2 years ago