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bagirrra123 [75]
3 years ago
14

A space shuttle can accelerate at a rate of 2.5m/s squared. Given that, and the fact that the engines of the space shuttle can p

roduce a force of about 1,600,000N, determine the mass(in kilograms) of the space shuttle
Physics
2 answers:
Afina-wow [57]3 years ago
8 0

Answer:

640000 kg

Explanation:

Formula is

Force = mass * acceleration

Mass = force / acceleration

Mass = 1600000 / 2.5

Mass = 640000 kg

ira [324]3 years ago
4 0
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maxonik [38]
Pressure=force/area
1.5/0.5=3Pa
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3 years ago
Read 2 more answers
an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?
gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

F_{net} = F_g - F = m\cdot g - F\\

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N

The tension force F in the supporting cables is 7245N


3 0
3 years ago
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar
Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>
8 0
1 year ago
Q3. You throw a ball into the air, it reaches a certain height and then comes back to you.
Grace [21]

The If a car is going round a curve , there is an acceleration because the direction of the velocity changes.

<h3>What is the direction of the velocity?</h3>

Now we know that if you throw the ball upwards, the motion is in opposite direction to gravity thus the ball is experiencing deceleration and the speed decreases. The velocity decreases and the acceleration is negative.

If the ball is coming down, then the ball is accelerated thus it speeds up and the direction of the acceleration is positive.

If a car is going round a curve, the vehicle is accelerating because the direction of the velocity changes even if its amount remains constant.

When a board is moving down a hill at 2 ms-1, it is experiencing an acceleration because the motion is in the same direction as gravity.

If a car is coming to a stop at a point, it experiences a deceleration and not an acceleration since the change of velocity with time is negative as the car comes to rest.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

5 0
1 year ago
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