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3 years ago

(DUE IN 1 HOUR) A rocket of mass m is launched straight up with thrust F⃗ thrust.

1 answer:

8 0

From the statement, since rocket was launched so this means that it start from zero, hence initial velocity is zero. 
Since the rocket was launched vertically straight up, the force acting on this motion is gravity.

A. The acceleration of the motion is then given by:

a= F/m - g

Then we can use the general equation:

V^2 = Vo^2 + 2*a*h

where V is final velocity, Vo is initial velocity, a is acceleration, h is height 

Since we know that Vo = 0, so

V^2 = 2*a*h

V^2 = 2 (F/m – g) h
V = sqrt [2 h (F/m – g)]

B. Given that:

h = 86 m

F = 10 N

m = 340 g = 0.34 kg

Find for V:

V = sqrt [2 * 86 (10 / 0.34 – 9.8)]

V = 58.08 m/s

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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

4 0

3 years ago

Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

$F_{x} + F{y} = ma$ (2)

The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

$F_{x} = 10 N - 4 N$

Notice that $f$ subtracts to $F$ since it is at the opposite direction.

$F_{x} = 6 N$

The forces present in y balance each other:

$F_{y} = 0$

Therefore:

$6 + 0 = ma$

$6 N = (2kg)a$ (3)

But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

So the block has an acceleration of $a = 3 m/s^{2}$ .

4 0

3 years ago

A box having a mass of 0.2 kg is dragged across a horizontal floor by

Lyrx [107]

you can check attachment for answer.

kind regards

7 0

3 years ago

Acceleration of gravity on jupiter

Viefleur [7K]

The acceleration of gravity on Jupiter is listed as 24.79 m/s² .

That's roughly 2.53 times its value on Earth. So if you weigh, let's say,
130 pounds on Earth, then you would weigh about 328 pounds on Jupiter.

5 0

4 years ago

Read 2 more answers

The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he

Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0

3 years ago