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Leya [2.2K]
3 years ago
7

The more ______ an object moves, the more __________ it has (Kinetic Energy)

Physics
1 answer:
belka [17]3 years ago
3 0

Answer:

__________

Explanation:

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Two trucks are traveling in the same direction, one going twice as fast as the other. at the end of 6 hours they are 204 miles a
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<span>speed of slower=x; distance in 6 hours is 6x
 speed of larger=2x; distance in 6 hours is 12x
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 x=34 mph slower, 204 miles in 6 hours.
 2x=68 mph, faster, 408 miles in 6 hours, and difference is 204 miles</span>
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Which element below has chemical properties similar to the chemical properties of fluorine.
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For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.
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3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
Joe the house painter stands on a uniform oak board weighing 600 N and held up by vertical ropes at each end. Joe has been dieti
Mamont248 [21]

Answer

given,

weight of the oak board = 600 N

Weight of Joe = 844 N

length of board = 4 m

Joe is standing at 1 m from left side

vertical wire is supporting at the end.

Assuming the system is in equilibrium

T₁ and T₂ be the tension at the ends of the wire

equating all the vertical force

T₁ + T₂ = 600 + 844

 T₁ + T₂ = 1444...........(1)

taking moment about T₂

 T₁ x 4 - 844 x 3 - 600 x 2 = 0

 T₁ x 4 = 3732

 T₁ = 933 N

from equation (1)

 T₂ = 1444 - 933

 T₂ = 511 N

3 0
4 years ago
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