Ask question

Ask question

All categories

3 years ago

8

A jumbo jet of mass 100,000 kg during takeoff experience a thrust for each of its four engines of 50,000 N. producing an acceler

ation of
2 m/s2
1 m/s
0.5 m/s2
O 4 m/s²
none of the above

1 answer:

PtichkaEL [24]3 years ago

8 0

Answer:

F=MA

5OOOO=100000×A

50000/100000=A

A=5/10 M/ S SQ

A=0.5 M/S SQ.

You might be interested in

At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp

motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

$K = \frac{1}{2} mv^2$

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51 = 1.07 kg

The mass of the body at that instant is 1.07 kg

5 0

3 years ago

The intensity of light from a star varies inversely as the square of the distance. If you lived on a planet ten times farther aw

frosja888 [35]

Answer:

the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.

That is,

Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100

Explanation:

Let the intensity of light be represented by I

Let the distance of the star be d

I ∝ (1/d²)

I = k/d²

For the earth,

Iₑ = k/dₑ²

k = Iₑdₑ²

For the other planet, let intensity be Iₒ and distance be dₒ

Iₒ = k/dₒ²

But dₒ = 10dₑ

Iₒ = k/(10dₑ)²

Iₒ = k/100dₑ²

But k = Iₑdₑ²

Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100

Iₒ = Iₑ/100

Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.

3 0

4 years ago

6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

5 0

3 years ago

Read 2 more answers

According to the kinetic molecular therory, the pressure of a gas in a container will increase if the

Masja [62]

According to the kinetic molecular theory, the pressure of a gas in a container will increase if the number of collisions with the container wall increases.

<u>Explanation:</u>

Keeping the volume of the vessel constant, if we increase the amount of gas in it; the pressure will increase. This is because when the number of gas particles increases in that limited volume, they hit the walls of the container with more energy and hence, the overall pressure of the gas increases.

If we decrease the amount of gas in the vessel or increase the volume for the same amount of gas, the pressure decreases. As the pressure inside the vessel depends upon the gas supplied in the container.

5 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago