Answer:
-4×-2y=14 (1)
-10×+7y=-25 (2)
multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2
-28×-14y=98
-20×+14y=-50
___________
-28×=48
×=48/-28
×=-12/7
now
-4×-2y=14
-4*-12/7-2y=14
48/7-2y=14
-2y=14-48/7
-2y=(98-48)/7
-2y=50/7
y=-50/14
y=-25/7
Answer: "B" Changing Position
Great Question!
Explanation: <u><em>When a ball bounces to the ground it hits the ground with some energy. The amount of energy with which it hits the ground is kinetic energy. When it comes in the contact with the ground kinetic energy gets converted into potential energy. This potential energy again gets converted into kinetic energy and balls moves again from the ground and bounces multiple times. So, the ball ends up changing position</em></u>
<u><em /></u>
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be 
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
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