Total distance covered is 47.1 m whereas displacement is zero.
<h3>Calculation:</h3>
Given,
Diameter, d = 5 m
No. of revolutions = 3
Radius, r = 5/2 = 2.5 m
To find,
Distance =?
Displacement =?
Distance covered in one revolution = 2πr
Put the values in this,
Distance = 2 × 3.14 × 2.5
= 15.7 m
Total distance covered in 3 revolution = 3 × 31.4
= 47.1 m
Displacement is the change in the position of the object or the distance between the initial and final position.
After 3 revolutions the particle comes back to its initial position. Therefore, the displacement is zero.
Hence, the total distance covered in 3 resolutions is 47.1 m whereas displacement is zero.
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If atoms lose , they become more organized.
When energy is removed from a given amount of atoms, they enter a more organized state because their attractive forces overcome their repulsive forces, which causes them to come closer. This is observed when a gas is condensed to a liquid, and then also when a liquid freezes into a solid. This moving closer also limits the motion of the particles.
The resistance of the conductor is 0.07940 Ohms.
<h3>What is the relation between resistance and area of wire?</h3>
The wire's resistance is inversely related to its cross-sectional area; as the area drops, so does the resistance.
and it is formulated as:
where,
<em>p </em>- resistivity of the conductor (0.0214-ohm mm²/m)
R - resistance
l- length of conductor (50 feet) (15.24 m)
s - the area of the wire
Thus the resistance can be calculated as
R = 0.07940 Ohms.
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Answer:
a.2.5x 10^3 m/s
b.mr=48kg/s
Explanation:
A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a speed of 5.0 ✕ 103 m/s relative to the rocket. The mass of the rocket at this time is 6.0 ✕ 104 kg, and its acceleration is 4.0 m/s2. What is the velocity of the exhaust relative to the solar system? (B) At what rate was the exhaust ejected during the firing?
velocity of the exhaust relative to the solar system
velocity of the rocket -velocity of the exhaust relative to the rocket.
7.5 ✕ 103 m/s-5.0 ✕ 103 m/s
2.5x 10^3 m/s
. b we will look for the thrust of the rocket
T=ma
T=6.0 ✕ 104 kg*4.0 m/s2
T=2.4*10^5N
f=mass rate *velocity of the exhaust
T=2.4*10^5N=mr*5.0 ✕ 10^3 m/s
mr=2.4*10^5N/5.0 ✕ 10^3
mr=48kg/s
Answer:
(a) the compression of the spring when the block comes to rest is 4cm
(b) speed of the block is 0.427 m/s
Explanation:
Given;
mass of the block, m = 1.6 kg
spring constant, k = 902 N/m
initial speed of the block, v₀ = 0.95 m/s
(a) the compression of the spring when the block comes to rest
when the block comes to rest, the final speed, vf = 0
Apply the law of conservation of energy;
¹/₂kx² = ¹/₂mv₀²
kx² = mv₀²
x = 4 cm
(b) speed of the block
Apply the law of conservation of energy;
¹/₂mv² = ¹/₂kx²
mv² = kx²
