A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a
2 answers:
Answer:
The mass of the second sphere is 15.4 kg
Explanation:
Given;
mass of the first sphere, m₁ = 7.7 kg
initial velocity of the second sphere, u₂ = 0
let mass of the second sphere = m₂
let the initial velocity of the first sphere = u₁
final velocity of the composite system, v = ¹/₃ x u₁ =
From the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
Substitute the given values;
Divide through by u₁
7.7 = ¹/₃(7.7 + m₂)
multiply both sides by 3
23.1 = 7.7 + m₂
m₂ = 23.1 - 7.7
m₂ = 15.4 kg
Therefore, the mass of the second sphere is 15.4 kg
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Answer:
a. The ratio of their resistance is 2783:64
b. The ratio of their energy is 4:23
c. The charge on the first bulb is 5.75 C
The charge on the second bulb is C
Explanation:
The voltage on one of the electric bulbs, V₁ = 40 volts
The power rating of the bulb, P₁ = 230 w
The voltage on the other electric bulbs, V₂ = 110 volts
The power rating of the bulb, P₂ = 40 w
a. The power is given by the formula, P = I·V = V²/R
Therefore, R = V²/P
For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96
The resistance of the second bulb, R₂ = 110²/40
The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64
∴ The ratio of their resistance, R₂:R₁ = 2783:64
b. The energy of a bulb, E = t × P
Where;
t = The time in which the bulb is powered on
∴ The energy of the first bulb, E₁ = 230 w × t
The energy of the second bulb, E₂ = 40 w × t
The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23
∴ The ratio of their energy, E₂:E₁ = 4:23
c. The charge on a bulb, 'Q', is given by the formula, Q = I × t
Where;
I = The current flowing through the bulb
From P = I·V, we get;
I = P/V
For the first bulb, the current, I = 230 w/40 V = 5.75 amperes
The charge on the first bulb per second (t = 1) is therefore;
Q₁ = 5.75 A × 1 s = 5.75 C
The charge on the first bulb, Q₁ = 5.75 C
Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = C
The charge on the second bulb, Q₂ = C.
d. The question has left out parts
Part of the question is missing. Here it is:
<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor? </em>
Answer:
Explanation:
The motion of the ball is a projectile motion, which consists of two separate motions:
- A horizontal motion at constant velocity
- A vertical motion at constant acceleration (free fall)
We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation
where, choosing downward as positive direction:
s =1.3 m is the vertical displacement of the ball
u = 0 is the initial vertical velocity
is the acceleration of gravity
t is the time
Solving for t,
Now we can find the final vertical velocity of the ball, using:
And susbtituting t = 0.52 s, we find
It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.
The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of
d = 0.7 m
In a time of
t = 0.52 s
So, the horizontal velocity is
So now we can find the direction of the ball's velocity using:
And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is