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3 years ago

Imagine that you drop a shot put from a tower on the moon. How much

Physics

1 answer:

ella [17]3 years ago

4 0

Answer: y = yo + Vyot + ayt = =2ay y− yo . 5. vy. 2. −voy. 2. =−2 g y− yo

Explanation: since both the displacement and acceleration are negative, they cancel each other when divided, so the result is positive. ... As you can see, as the height displacement (height) increases, the longer it takes for the robocopter to fall.

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The first mechanized industry was

Oxana [17]

C. Textiles

It was the first thing mechanized in the Industrial Revolution

3 0

3 years ago

How would I workout question 4?? Please help

Agata [3.3K]

Answer:

550 kg

Explanation:

mass = E / gh

= 33000/60

=550

plzzz......

mark it as a brilliant answer

3 0

3 years ago

Giving brainliest!! if you answer correctly :) (30pts)

AnnZ [28]

Answer:

32000joule.

Explanation:

given

mass. (m)=160kg

speed (v)=20m/s

now

kinetic energy =1/2 (mv²)=1/2 ×{160×20²}=32000joule.

8 0

3 years ago

Read 2 more answers

How do magnetic forces repel or attract?

Lerok [7]

“Magnets are surrounded by an invisible magnetic field that is made by the movement of electrons, the subatomic particles that circle the nucleus of an atom”

“Every magnet has both a north and a south pole. When you place the north pole of one magnet near the south pole of another magnet, they are attracted to one another. When you place like poles of two magnets near each other (north to north or south to south), they will repel each other.”

3 0

3 years ago

A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is

Allushta [10]

Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

$Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }$

$Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }$

Equaling both Reynold's number:

$\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }$

Clearing Vm:

$V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s$

b) The drag force is:

$\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N$

6 0

3 years ago

Read 2 more answers