A cord is attached to the box and run through a pulley directly above the box, so that the cord is vertical. The free end of the
cord is then connected to a hanging weight. Calculate the force exerted by the ground on the box (in lb), if the hanging weight is 10.5 lb.
1 answer:
Answer:
The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.
<u>F1 = T- 46, 674.656 gm/s² </u>
Explanation:
<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>
To calculate the force caused by gravity on the basic pulley system we use the following equation:
F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g
∴ F2 = 4762.72g x 9.8 m/s²
= 46, 674.656 gm/s² or 46, 674.656 N
But since this F2 is acting in a downlowrd direction, it would be negative.
Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)
⇒ F1 = T - F2
<u>F1 = T- 46, 674.656 gm/s² </u>
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Answer:
<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>
Explanation:
<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.
Here it is given that the object oscillates 20 times in 10 seconds.
So f = = 2Hz
The <em>time period</em> is defined as time taken by the object to complete one full oscillation.
T =
T= =0.5 s
<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>
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