Answer:
4.15 m/s
Explanation:
Its given that acceleration is 0.1 m/s² with a direction opposite to the velocity. Since, the direction of acceleration is opposite to the velocity, this gives us a hint that the velocity is decreasing and so acceleration would be negative.
i.e.
acceleration = a = - 0.1 m/s²
Distance covered = S = 6m
Velocity after covering 6 meters = Final velocity =
= 4 m/s
We need to find the initial speed, which will be the same as the magnitude of initial velocity.
Initial velocity =
= ?
3rd equation of motion relates the acceleration, distance, final velocity and initial velocity as:
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Using the known values in the formula, we get:

Thus, the initial speed of the ball was 4.15 m/s
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Answer:
2N
Explanation:
subtract rthe two forces to see which is greater
4-2=2
Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.
According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J
Minus sign indicates that work is done by the system.
Answer:
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.
Explanation:
Mass of object 1 , m₁ = 300 g = 0.3 kg
Mass of object 2 , m₂ = 400 g = 0.4 kg
Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s
Initial velocity of object 2 , v₂ = 3.00j m/s
Mass of composite = 0.7 kg
We need to find final velocity of composite.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s
Final momentum = 0.7 x v = 0.7v kgm/s
Comparing
1.5 i + 0.24 j = 0.7v
v = 2.14 i + 0.34 j
Magnitude of velocity

Direction,

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.