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3 years ago

Can an object have a low acceleration and high velocity at the same time . give example

2 answers:

5 0

Yes, acceleration only tells you how velocity is changing. It doesn't say anything about what velocity is at any given time.

For example, if you set your car to cruise control on the highway going 80 mph. That is a constant high velocity, yet the car has 0 acceleration.

The opposite is also true. After a red light turns green, you put foot on the gas to accelerate. However, your velocity is initially low even though it has high acceleration.

Elis [28]3 years ago

4 0

Of course !

A passenger jet, cruising at 37,000 feet on its way to California,
keeping a constant airspeed and flying in a straight line, can have
a velocity of 500 miles per hour west, but its acceleration is zero !

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3 years ago

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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

marshall27 [118]

Answer:

The resultant velocity is $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

$M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

$v_L$ is the velocity of Luxury Liner = 20 knots due west

$M_f$ mass of freighter = 60,000

$v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

$v_f = 0 \ knots$

So the equation would be

$M_L *v_L = (M_L + M_f) v_t$

Substituting values

$40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

$v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

$= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction

$v_L = 0 \ knots$

So the equation would be

$M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

Making $v_t_n$ the subject

$v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

$= 6 \ knots$

The resultant velocity is

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$v_t=10 knots$

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2 years ago

if an object is thrown straight up with an initial velocity of 8m/s and takes 3 seconds to strike the ground, from what height w

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Answer:

s = it+1/2 at²

s= 8×3+1/2 (10)(3)²

s = 24+45

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2 years ago

What is the process called when the moon begins to fade from a full moon to the new moon?

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Answer: Waning

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3 years ago

In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas

Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

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3 years ago

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