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3 years ago

Heyyy! Okay riddle of the day

Physics

1 answer:

vekshin13 years ago

3 0

Answer:

Time to get a new fence...?

C'mon that's pretty cringe lol

Please give brainliest

have a nice day :D

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What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Earth-atmospheres.

EleoNora [17]

Below is an attachment containing the solution.

4 0

3 years ago

Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constan

elena55 [62]

Answer:

$10.07m/s^2$

Explanation:

Information we have:

velocities:

initial velocity: $v_{i}=0mph$ (starts from rest)

final velocity: $v_{f}=50mph$

time: $t=2.22s$

Since we need the answer in $m/s^2$ , we nees to convert the speed to meters per second:

$v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s$

We find the acceleration with the following formula:

$a=\frac{v_{f}-v_{i}}{t}$

substituting the known values:

$a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2$

the acceleration is 10.07 $m/s^2$

8 0

3 years ago

Read 2 more answers

HELP ME PLS!!!!

alexgriva [62]

Answer:

C. An ion with a +2 charge

Explanation:

I guess

8 0

2 years ago

Read 2 more answers

Why is water pollution a concern if water is continuously cycled through Earth’s systems?

kati45 [8]

Answer:Water pollution is a concern because since it is cycled threw the earths system it effect the earths system in a negative way.

Explanation:

I would wait for someone else to answer just in case I’m wrong.

7 0

3 years ago

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0

3 years ago