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2 years ago

10

In a scientific test conducted in Arizona, a special cannon called HARP (High Altitude Research Project) shot a projectile strai

ght up to an alti-tude of 180.0 km. If the projectile's initial speed was 3.00 km/s, how long did it take the projectile to reach its maximum height?

1 answer:

Alexus [3.1K]2 years ago

5 0

Answer:

It took the projectile 120 s to reach the maximum height.

Explanation:

Given;

maximum height of the projectile, s = 180 km = 180,000 m

initial speed of the projectile, u = 3 km/s = 3000 m/s

final velocity at maximum height, v = 0

Apply the following kinematic equation for average velocity of the projectile;

$s = (\frac{v+u}{2} )t\\\\(v+u)t = 2s\\\\t = \frac{2s}{v+u} \\\\t = \frac{2*180,000}{0+3,000}\\\\ t = 120 \ s$

Therefore, it took the projectile 120 s to reach the maximum height.

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Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the frequency of the waves is doubled

m_a_m_a [10]

Answer:

halved

Explanation:

The velocity of the a wave is obtained by multiplying the frequency and wavelength.

$v=f\lambda\\\Rightarrow f=\frac{v}{\lambda}\\\Rightarrow \lambda=\frac{v}{f}$

Where

v = Velocity

f = Frequency

$\lambda$ = Wavelength

The velocity here is constant. So, if the frequency is doubled the wavelength is halved.

6 0

2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

6 0

1 year ago

Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor

grin007 [14]

The job that the fan is designed and built to do is to convert the electrical energy it uses into the kinetic (motion) energy of moving air.

I can't really guarantee that it accomplishes that with MOST of the electrical energy it uses, because I don't know how efficient your fan is. For example, if it's a really old fan, and one blade has the end broken off, and a lot of dust and mosquitoes have gotten into the motor, and it shakes and vibrates and makes a lot of noise when it's running, then it's converting a lot of the electrical energy into thermal energy (it gets hot when it runs) and some into sound energy too.

If you can live without the word "most" in the question, then we can assume that the fan is well designed and running like a top, and the answer is definitely choice-B .

4 0

2 years ago

It takes 300 newtons of force and a distance of 20 meters for a moving car to come to stop

Arlecino [84]

The answer is 15 I think.

6 0

2 years ago

You jump off a platform 134 m above the Neuse River. After you have free-fallen for the first 40 m, the bungee cord attached to

omeli [17]

Answer:

The acceleration at lowest point is 19.62 m/s^2

Explanation:

Conservation of energy is an concept in which it is stated that the energy of an isolated object remains the same. Energy changes from one form to another.

Lets Assume

Constant of string is K

By using the conservation of energy we will have the following equation

1/2 x 80^2 x K = m x 9.81 x 120

3200 K = 1177.2 m

K = 1177.2 m / 3200

K = 0.368 m

At the lowest point we will have

a = ( K x X - m x g ) / m

a = ( 0.368 m x 80 - m x 9.81 ) / m

a = 19.62 m / s^2

So, the acceleration at lowest point is 19.62 m/s^2

7 0

2 years ago

Read 2 more answers