Indicate the substance that is NOT an element.

Find the equation of the line passing through the points(-3,0)and(2,-3).<br>asap

AleksandrR [38]

This isn’t physics assnugget

7 0

3 years ago

How far (in meters) will you travel in 3 minutes running at a rate of m/s?

Daniel [21]

10 minutes I don’t know id this is right

3 0

3 years ago

Read 2 more answers

A spaceship, at rest in some inertial frame in space, suddenly needs to accelerate. The ship forcibly expels 103 kg of fuel from

Cerrena [4.2K]

Answer:

$V_s = 1.8*10^5m/s$

Explanation:

There is no external force applied, therefore there is a moment's preservation throughout the trajectory.

<em>Initial momentum = Final momentum. </em>

The total mass is equal to

$m_T= m_1 +m_2$

Where,

$m_1 =$ mass of ship

$m_2 =$ mass of fuell expeled.

As the moment is conserved we have,

$0=V_fm_2+V_sm_1$

Where,

$V_f =$ Velocity of fuel

$V_s =$ Velocity of Space Ship

Solving and re-arrange to $V_s$ we have,

$V_s = \frac{V_f m_2 }{m_1}$

$V_s = \frac{3/5c}{10^6}$

$V_s = 3.5*10^{-3}c$

Where c is the speed of light.

Therefore the ship be moving with speed

$V_s = \frac{3}{5}*10^{-3}*3*10^8m/s$

$V_s = 1.8*10^5m/s$

5 0

3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

3 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago