All categories

3 years ago

A. An automobile mass is 3.5 x103 kg. If the forward thrust (Fnet)

Physics

1 answer:

katrin [286]3 years ago

8 0

Answer:

a = 0.8 m/s^2

Explanation:

Force equation: F = ma

F = ma -> a = F/m = 2.8*10^3 N / 3.5*10^3 kg = 0.8 m/s^2

You might be interested in

An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that

SIZIF [17.4K]

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

6 0

3 years ago

Read 2 more answers

wo slightly out of tune instruments will create beats when the same note is played. This is a pattern of louder and softer sound

Lostsunrise [7]

This interaction is known as constructive interference. It's a result of linear superposition.

5 0

3 years ago

2. What is one way that science helps shape society?

zhuklara [117]

Answer:Science is valued by society because the application of scientific knowledge helps to satisfy many basic human needs and improve living standards. Finding a cure for cancer and a clean form of energy are just two topical examples.

Explanation:

4 0

3 years ago

How much force is needed to keep the 750000 Newton Space Shuttle moving at a constant speed of 28000 km/h, in a straight line?

Alika [10]

The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

weight of the space shuttle, F = 750,000 N
constant speed of the space shuttle, v = 28,000 km/h

The mass of the space shuttle is calculated as follows;

$W = mg\\\\m = \frac{W}{g} \\\\m = \frac{750,000}{9.8} \\\\m = 76,530.61 \ kg$

The force needed to keep the space shuttle moving at constant speed is calculated as follows;

$F = ma$

$F = 76,530.61 \times a$

where;

a is the acceleration of the space shuttle

At a constant speed, acceleration is zero.

F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

Learn more here:brainly.com/question/16374764

6 0

2 years ago

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago