The kinetic energy of a 1.0 kg ball thrown with an initial velocity of 30 m/s is calculated using the formula KE = 1/2 mv^2, resulting in an energy of 450 Joules.
The question you've asked pertains to calculating the kinetic energy of a ball thrown into the air. To find the kinetic energy (KE) of a 1.0 kg ball thrown with an initial velocity of 30 m/s, you can use the formula KE = ½ mv², where m is the mass of the ball and v is the velocity. Plugging in the values, you get KE = ½ × 1.0 kg × (30 m/s)² = 0.5 × 1.0 × 900 = 450 J. Therefore, the kinetic energy of the ball is 450 Joules.
#SPJ12
Explanation :
The resistance of the flashlight is 2.4 ohms
The current flowing in the circuit is 2.5 A
The Ohm's law gives the relation between the following quantities i.e.
Current
voltage
and resistance
Mathematically, Ohm's law can be written as :
V =I R
The voltage applied by the batteries is 6 volts.
Hence, this is the required solution.
Answer:
D - 6.0
Explanation:
i'm not getting more detailed than the other answer so yeah i just took the test
b) False
Answer:
b)False
Explanation:
This is the false statement .The two forces in each pair could not have different physical origins.
The two pair of forces have same origin.It can not be possible that one is electric force and other one is gravity force or friction force.
We know that magnitude of electric force is much smaller than the friction force .So they can not be in pair form.
Answer is b) False
Answer:
4.281 kgm/s upward
Explanation:
Impulse:This can be defined product of force and time. The S.I unit of impulse is Ns.
From Newton's second law of motion,
Impulse = Change in momentum.
I = mΔv....................... Equation 1.
Where m = mass of the ball, Δv = change in velocity of the ball
and Δv = v -u
Where u = velocity of the ball before it hit the floor, v = velocity of the ball after if hit the floor
I = m(v-u) -------------- Equation 2
But
the initial kinetic energy of the ball = potential energy at the initial height (1.2 m above)
1/2mu² = mgh₁
Where h₁ = initial height. or height of the ball before collision
making u the subject of the equation,
u = √(2gh₁)........................ Equation 3
Where h₁ = 1.2 m g = 9.81 m/s²
Substitute into equation 3
u = √(2×1.2×9.81)
u =√(23.544)
u = -4.852 m/s.
Note: u is negative because the ball was moving downward at the first instance.
Similarly,
v = √(2gh₂)............................. Equation 3
h₂ = height of the ball after collision
Given: h₂ = 0.7 m, g = 9.81 m/s²
Substitute into equation
v = √(2×9.81×0.7)
v = √13.734
v = 3.71 m/s.
Also given: m = 0.5 kg,
Substituting into equation 2
I = 0.5(3.71-(4.852)
I = 0.5(8.562)
I = 4.281 kgm/s. Upward.
Thus the impulse = 4.281 kgm/s upward