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Sliva [168]
3 years ago
10

WILL GIVE BRAINLIEST PLEASE I NEED THIS! A 100-kg projectile crashes through a castle wall with a kinetic energy of 266.45 kJ. W

hat was the speed of the projectile?
Physics
1 answer:
denpristay [2]3 years ago
7 0
K.E.=
\frac{1}{2} m {v}^{2}
so just sub in the info you are provided with.
266.45x10^3=0.5×100xv^2
V is 73.
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What's the value of 1,152 Btu in joules? A. 1,964,445 J B. 2,485,664 J C. 987,875 J D. 1,215,360 J
OLga [1]

Answer: Option (d) is correct.

Explanation:

Given,  1,152 British thermal units

1 British thermal unit = 1055.06 joules

So, in 1,152 British thermal units there will be :

1152\times 1055.06 Joules=1215429.12 Joules=1.21542912\times 10^{8} Joules

Hence, from the given options the closest answer is of option (d). So, option (d) is correct.

4 0
3 years ago
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How much heat is needed to melt 16.5kg of silver that is initially at 20*c?
zubka84 [21]
The Specific Heat Capacity of silver is 230 J/kgK, melting point is 961.8 C so the difference is 941.8K. Now we simply do q=230J/kgK*16.5kg*941.8K and that is 3 574 131 J
8 0
3 years ago
What will be the value of both charges if they are 5 cm apart and suffer a
Mrac [35]

Answer:

\boxed{q = 1.2 \times  {10}^{ - 6} C}

Explanation:

f_e =   \frac{{q}^{2}k }{ {r}^{2} }  \\ q =  \sqrt{ \frac{f_e( {r}^{2} )}{k} }  =  \sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  \\ q =\sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  =  \sqrt{ \frac{0.013}{9 \times  {10}^{9} } }  \\ q = 1.2 \times  {10}^{ - 6}

3 0
2 years ago
In order for a space shuttle to leave Earth, it must produce a great amount of thrust. Its rocket boosters create this thrusting
bonufazy [111]

Answer:

Orbit

Explanation:

The term that describes motion of the shuttle around earth is called as<em> Orbit.</em>

The orbit is defined as a regular repeating path that object takes around another.

The shuttle circles around the earth at a constant distance from earth surface is because of earth gravity and forward motion of earth.

5 0
3 years ago
A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point
Mandarinka [93]

Answer:

F=9.09\times 10^{-9} N

Explanation:

We are given that

Mass of spherical shell,m_1=1900 kg

Mass=m_2=1.80 kg

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

F=G\frac{m_1m_2}{r^2}

G=6.67\times 10^{-11} Nm^2/kg^2

Using the formula

F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}

F=9.09\times 10^{-9} N

Hence,the gravitational force =F=9.09\times 10^{-9} N

6 0
3 years ago
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