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barxatty [35]
3 years ago
6

Day and night are caused by

Physics
1 answer:
JulijaS [17]3 years ago
4 0

Answer:

A

Explanation:

How is the rotation of the earth responsible for causing day and night?

We get day and night because the Earth spins (or rotates) on an imaginary line called its axis and different parts of the planet are facing towards the Sun or away from it. It takes 24 hours for the world to turn all the way around, and we call this a day.

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A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat
jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

I=Ft

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

I=Ft=(40 N)(3.0 s)=120 Ns

8 0
3 years ago
I just need the answer
Archy [21]

Answer:

I'll try to help which grade are you?

7 0
3 years ago
The flow rate in a firehose is 0.524 m3/s. It is able to shoot water to the top of a building 40.4 m tall, but not higher. You r
Marrrta [24]

Answer:

A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.

Explanation:

3 0
3 years ago
Look at the velocity versus time graph below. What is the magnitude of the displacement of the object after it travels for three
Elza [17]

Answer:

C. 12m

Explanation:

veocity =  \frac{displacement}{time}

from the graph v = 4m/s and t = 3 s

d = vt = 4 × 3 = 12 m

5 0
3 years ago
An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
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