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2 years ago

2. A rocket is launched with an upward acceleration of 20.0 m/s2 that lasts for 4.50 s. During

Physics

1 answer:

Neporo4naja [7]2 years ago

8 0

Answer:

Ggghghhh

Explanation:

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A skateboader is accelerating forward how would his acceleration change if he had more mass

Masteriza [31]

Answer: If the force stays the same, the acceleration would decrease

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2 years ago

Problem: Suppose that the rate of flow of water through a fire hose is 21.4 kg/s and the stream of water from the hose moves at

Sloan [31]

Answer:

The force exerted is 318.86 N

Explanation:

The force exerted by such a stream is calculated by multiplying the mass flow rate of water by the velocity of the stream of water.

mass flow rate = 21.4 kg/s

velocity = 14.9 m/s

Force exerted = 21.4 kg/s × 14.9 m/s = 318.86 kgm/s^2 = 318.86 N

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3 years ago

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With a velocity of 15 m/s, how long would it take the dog to run 100 meters?

Zinaida [17]

Answer

6.66 seconds imma seem like an idiot if something like this is werong

Explanation:

dogg is speed

100/15

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2 years ago

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a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra

Feliz [49]

Answer:

0.331m

Explanation:

wave equation: v=f λ

v=30000m/s

f=90500 Hz

λ= $\frac{v}{f}$

λ= $\frac{30000}{90500}$

λ= 0.311m

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1 year ago

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

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3 years ago