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pickupchik [31]

4 years ago

8

In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other e

nd rests on a concrete pillar. When a clown (mass 79.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical. Calculate the force exerted by the rope on the board with the clown on it.

1 answer:

tester [92]4 years ago

4 0

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

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4 years ago

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential

kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$ .

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

<h3>Part A</h3>

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$ . The voltage across the two resistor, combined, is equal to $\rm 6\; V$ . Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$ .

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$ . Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$ .

<h3>Part B</h3>

Since the equivalent resistance is equal to $R_1 + R_2$ , when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$ . When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

<h3>Part C</h3>

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$I(R_1) = I(\text{Circuit}) = I(R_2)$ .

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

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4 0

3 years ago

a sphere with a radius of 8cm carries a uniform volume charge density of 1.5 find the magnitude of the electric field

gladu [14]

Answer:

E = 5.65 x 10¹⁰ N/C

Explanation:

First we need to find the total charge on the sphere. So, we use the following formula for that purpose:

$q = \sigma V\\$

where,

q = total charge on sphere

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Therefore,

$q = (0.335\ m^3)(1.5\ C/m^3) \\q = 0.502 C$

Now, we use the following formula to find the electric field due to this charged sphere:

$E = \frac{kq}{r^2}$

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E = Electric Field Magnitude = ?

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r = radius of sphere = 8 cm = 0.08 m

Therefore,

$E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(0.502 C)}{(0.08\ m)^2}\\\\$

<u>E = 5.65 x 10¹⁰ N/C</u>

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