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3 years ago

An object is given an initial velocity. What will happen to the object if no other forces act on it?

Physics

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

The object will travel at a constant rate in along a straight line.

Explanation:

In the given situation, it is mentioned that there is no external force acting on the given object. Thus, it will retain its initial velocity along a straight path.

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Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

vichka [17]

It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>

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3 years ago

Read 2 more answers

A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the

Dvinal [7]

Displacement = distance and direction from the start-point
to the end-point, regardless of the route followed on the way.

From the throw to the 'plop', the displacement is 5 meters down.

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3 years ago

Find the density of seawater at a depth where the pressure is 680 atm if the density at the surface is 1030 kg/m3. Seawater has

Pie

Answer:

$1060.41kg/m^3$

Explanation:

Bulk modulus is defined as the relative change in the volume of a body produced by a unit of compressive acting uniformly over its surface:

$B=\rho _o \frac{\bigtriangleup P }{\bigtriangleup \rho}$

Hence the density of the seawater at a depth of 680atm is calculated as:-

$\rho=\rho_o +\bigtriangleup \rho=\rho_o(1+\frac{\bigtriangleup P}{B})\\\\=1030 \times (1+ \frac{(680-1)\times10^5}{2.3\times 10^9})\\=1060.41kg/m^3$

4 0

3 years ago

A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per

FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0

3 years ago

A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel

Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0

3 years ago