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3 years ago

11

A 1,500-kg car accelerates along a flat track to a speed of 20 m/s. how much does it's gravitational potential energy increases

over the length of the track?

1 answer:

lesya [120]3 years ago

3 0

Look at bitesizes physics section, they have all the information you need to complete this question.

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A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

3 years ago

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0

3 years ago

(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e

gtnhenbr [62]

Answer:

$C=-\dfrac{10}{17}(F-32)$

Explanation:

Given that

32° F corresponds to 0 °C. ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

$y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)$

Lets C in y- direction and F in x- direction,so we can say that

$C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)$

$C-0=\dfrac{100-0}{32-212}(F-32)$

$C=-\dfrac{10}{17}(F-32)$

So the linear relationship is

$C=-\dfrac{10}{17}(F-32)$

8 0

3 years ago

Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer

sveticcg [70]

by cosine law we know that

$c^2 = a^2 + b^2 - 2 abcos\theta$

$\theta = 180 - 54.3 - 79.1 = 46.6 degree$

now using above equation

$93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6$

$4942.03 = b^2 - 85.34 b$

$b^2 - 85.34b - 4942.03 = 0$

by solving above quadratic equation we have

$b = 124.9 m$

so it is at distance 124.9 m from deer a

4 0

3 years ago

Why can’t we see black holes? Question 11 options: A.Because they don't want to be seen. B.Because the sun is too far away from

Vika [28.1K]

We can't see black holes because D) no light can get out

Explanation:

Black holes are the result of the gravitational collapse of a supermassive star.

The life of a supermassive star ends with a huge explosion, called supernova, that leaves behing a super-dense core called black hole.

Black holes are the most dense objects of the universe, having a huge mass in a super small size. For this reason, the gravitational force exerted by a black hole in its proximity is so strong that even light is not able to escape from the gravitational field. For this reason, light from a black hole is not able to reach us, and so we are not able to see these objects.

The "edge" of space beyond which light remains "trapped" inside the black hole is called event horizon, and no object can escape this region of space.

The radius of the event horizon of a black hole is called Schwarzschild radius and it is given by:

$r=\frac{2GM}{c^2}$

where

G is the gravitational constant

M is the mass of the black hole

c is the speed of light

Learn more about space:

brainly.com/question/2887352

brainly.com/question/10934170

#LearnwithBrainly

6 0

3 years ago