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3 years ago

9

Que formas de energía utilizas habitualmente?

1 answer:

Luda [366]3 years ago

5 0

Esta energía<span> puede ser convertida en otras, como calor para calentar agua o edificios, invernaderos etc. o electricidad. Podemos convertir la </span>energía<span> solar en eléctrica de dos </span>formas<span>: Fotovoltáica (PV): La radiación solar se convierte directamente en electricidad

hope this help mark brainliest plz</span>

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What are 2 ways organisms would be affected by water pollution?

Ymorist [56]

Not having clean water to drink or bath. Also some organizations that live in the water might die because the water is polluted and that would be toxic form them

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3 years ago

What is the acceleration of a car going 50 mph that slows down to rest over 10 seconds?

VladimirAG [237]

Answer:

If the velocity is constant, then there is no acceleration. That is, the value of the acceleration is 0.

Explanation:

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3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

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3 years ago

What can be found in a compost bin?

BigorU [14]

D. all of the above

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3 years ago

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