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MatroZZZ [7]
3 years ago
5

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)

Physics
2 answers:
anyanavicka [17]3 years ago
8 0

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

mgh=360800J

m×9.8×25=360800

m=360800/(9.8×25)=1472.653061kg

Archy [21]3 years ago
8 0

Answer:

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted!

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Waht scientific name of a horse is Equus callabux what is the genus name of the horse
Zarrin [17]

Answer:

middle

Explanation:

4 0
2 years ago
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

work done by gravity+work done by friction=Energy stored in Spring

mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

6 0
3 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c
Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

4 0
3 years ago
Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat
Nataly_w [17]

Answer:

\omega_{f} = 1000000\,\frac{rev}{day}

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}

The final angular speed is:

\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}

\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}

\omega_{f} = 1000000\,\frac{rev}{day}

3 0
2 years ago
If you wanted to increase the gravitational force between two objects what would you do
adelina 88 [10]
Double the force on the object
6 0
3 years ago
Read 2 more answers
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