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3 years ago

A handel or knob is fixed at the free end of door. Why?

1 answer:

5 0

Maximum torque.
T=F.r
r is measured from point of turning, which is the hinge. r is maximum when the handle is at the free edge.

You might be interested in

A boat moves with a constant speed, covering 20.0m in a time of 4.0s

san4es73 [151]

Answer:

velocity = 5m/s

initial position : 0m

final position : 20 m

time = 4 s

5 0

3 years ago

A sports car moving at constant speed travels 164 m in 13.77 s. If it then brakes and comes to a stop in 3.6 s. What is its acce

Sergio039 [100]

Answer:

a = - 0.3376 g's

Explanation:

The sports car has a constant speed when travelling. Covered 164 m in 13.77 s. Thus, speed = 164/13.77 m/s

It brakes and now comes to a stop in 3.6 s.

Thus final velocity = 0 m/s

Formula for acceleration is;

a = (v - u)/t

a = (0 - (164/13.77))/3.6

a = -3.308 m/s²

In terms of g's", where 1.00 g = 9.80 m/s², we have;

a = -3.308/9.8 g's

a = - 0.3376 g's

4 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

3 years ago

Do humans have a gravitational field if so, how strong

lilavasa [31]

I'm pretty sure that you have mass. When it comes to gravity, anything that has mass behaves exactly like anything else that has mass. Any two of them are attracted to each other, with a force that's proportional to the product of their masses, and inversely proportional to the square of the distance between them.

Now, if you have mass, then you have weight ... it's the answer you give when somebody asks "How much do you weigh ?". That's the force of gravity between you and the Earth, pulling you toward the center of the Earth. But it doesn't stop there. There's also a force of gravity pulling the Earth toward the center of you. The strength of it is EQUAL to your weight.

Your weight on Earth is EQUAL to the Earth's weight on YOU !

7 0

3 years ago

Read 2 more answers

Choose the best description of the attraction between you and your classmate.

Lerok [7]

Answer:

The correct option is;

D. There is not enough information to answer this question

Explanation:

The universal gravitational constant = 6.67408 × 10⁻¹¹ 3³/(kg·s²)

For an in between distance of 1 m and equal masses of 60 kg, we have;

$F = G \times \dfrac{m_{1} \times m_{2}}{r^{2}} = 6.67408 \times 10^{-11} \times \dfrac{60 \times 60}{1^{2}} \approx 2.403 \times 10^{-7} N$

The gravitational attraction ≈ 2.403 × 10⁻⁷ N, which does not correspond with the answers, therefore, the best option is that there is not enough information to answer this question.

5 0

3 years ago