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3 years ago

A model rocket weighing 4.9 N launches from a pad. How much work does the little rocket engine do in the first 4 meters

Physics

1 answer:

USPshnik [31]3 years ago

5 0

Answer:

Work done by the little rocket engine = 19.6 Joules.

Explanation:

We know that work = force x displacement.

Force = weight of rocket = 4.9 N

Displacement = 5 meter.

So work done by the little rocket engine = Weight of rocket x displacement.

Work done by the little rocket engine = 4.9 x 4 = 19.6 J

Work done by the little rocket engine = 19.6 Joules.

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supposed the Kingfisher Dives with an average speed of 4.6 m / s for 1.4 seconds before hitting the water. What is the height fr

timama [110]

Answer:

h = 1.07 m

Explanation:

It is given that,

The average sped of the kingfisher, v = 4.6 m/s

Time, t = 1.4 s

Final speed, v = 0 (as it hits the water)

We need to find the height from which the bird dove. Let the height is h. According to the conservation of energy.

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(4.6)^2}{2\times 9.8}\\\\h=1.07\ m$

So, the bird dove from a height of 1.07 m.

3 0

4 years ago

According to American Heart Association, your target heart rate is__________.

sladkih [1.3K]

Answer:

Explanation:

A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal

4 0

3 years ago

A force of 31 N south is used to accelerate an object uniformly from rest to 8 m/s in 22 s. What is the mass of the object?

harina [27]

this question is centered around newtons second law f=ma

First you need to find the acceleration

$acceleration = \frac{8 - 0}{22}$

that gives you 0.36 recurring(on both numbers) m/s²

mass= force ÷ acceleration

$mass = \frac{31}{0.36recurring}$

mass= 85.25 kg

if you want to check

f=ma

85.25 × 0.36 recurring (remember its on both numbers not just 6) = 31 N

4 0

3 years ago

Read 2 more answers

Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x

GalinKa [24]

Answer:

a) (-367231.63i , 367231.63i, 0) N/C

b) (0 , 0 , 367231.63i ) N/C

Explanation:

Case x < -2.15

$E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i$

Case x > 2.15

$E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i$

Case -2.15 < x <+2.15

$E =E_{+} - E_{+} \\\\E = 0$

Case x < -2.15

$E =E_{+} - E_{+} \\\\E = 0$

Case x > 2.15

$E =E_{+} - E_{+} \\\\E = 0$

Case -2.15 < x <+2.15

$E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i$

7 0

3 years ago

A ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second2. What is the final velocity of the

cupoosta [38]

19.2 meters/second^2 would be the correct answer.

Try multiplying 3.2 meters/second^2 by 6 and you will receive the answer provided above. If you have any further questions, let me know!

8 0

3 years ago

Read 2 more answers