Answer:
Vf=60m/s
Vi=0 because it starts from rest
t=4.4s
a=?
a=Vf-Vi/t
a=60-0/4.4
a=60/4.4
a=13.63m/s^2
Answer:
13.636364 m/s2
B. resistance
C. potential difference
D. induction
The kinetic energy of the object with a mass of 2kg has a linear momentum of 6kgm/s is 9 joules.
Kinetic energy is the energy obtained when the body is in motion. It is obtained from the product of mass and velocity. The SI unit of Kinetic energy is Joule(J).
K.E = 1/2 (mv²) where m is the mass of the object and its unit is the kilogram. v is the velocity of the object and its unit is m/s. The momentum of the object is obtained from the product of mass and velocity.
From the given,
Mass of the object (m) = 2 kg
Linear momentum of the object = 6 kg.m/s
Linear momentum (p) = m×v (m = mass of the object, v= velocity of the object)
v = p / m
= 6 / 2
= 3 m/s
The velocity of the object = 3 m/s
Kinetic energy = 1/2 (mv²)
K.E = 1/2 (2×3×3)
= 9 joules
Thus the kinetic energy of the object is 9 joules (J).
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The impact of the heavy stone would produce with a higher amplitude than the impact of the light stone would because of having high potential energy.
An object's gravitational potential energy is described as the energy it would have had if it had defied gravity and risen to a certain height. Due to having only magnitude and no direction, it is a scalar quantity with SI unit of Joule.
Let, the mass of the heavy stone is m₁ and mass of the light stone is m₂ (m₁ > m₂). Both of them are dropped from same height h.
Then, gravitational potential energy of heavy stone is, V₁ = m₁gh.
And, gravitational potential energy of lighter stone is, V₂ = m₂gh.
Where, g = acceleration due to gravity.
Now, as m₁ > m₂; V₁> V₂
Due to having more potential energy, when the stones are dropped into a pond from same height, the heavy stone would produce with a higher amplitude than the impact of the light stone would.
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Answer:
The temperature of the Aluminium plate 44.84⁰C
Explanation:
Number of transistors = 4
Since the heat dissipated by each transistor is 12W
Total heat dissipated, Q = 4 * 12 = 48 W
Q = 48 W
Cross sectional Area of the Aluminium plate, A = 2(l * b)
l = Length of the aluminium plate = 22 cm = 0.22 m
b = width of the aluminium plate = 22 cm = 0.22 m
A =2( 0.22 * 0.22 )
A = 0.0968 m²
From the heat balance equation, Q = hAΔT
h = 25 W/m²·K
A = 0.0968 m²
ΔT = T - T(air)
T(air) = 25°C
ΔT = T - 25°C
Q = 25 * 0.0968 * ( T - 25)
Q = 2.42 (T - 25)
Substitute Q = 48 into the equation above
48 = 2.42 (T - 25)
T - 25 = 19.84
T = 25 + 19.84
T = 44.84 ⁰C