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3 years ago

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A block with mass m = 4.3 kg is attached to two springs with spring constants kleft = 35 N/m and kright = 48 N/m. The block is p

ulled a distance x = 0.23 m to the left of its equilibrium position and released from rest.What is the magnitude of the net force on the block (the moment it is released)?

1 answer:

Mrrafil [7]3 years ago

4 0

Answer:

Explanation:

A restoring force is created when a spring is either stretched or compressed.

Restoring force created by first spring = spring constant x contraction

= 35 x .23

= 8.05 N .

It will act towards the right .

Restoring force created by second spring = spring constant x extension

= 48 x .23

= 11.04 N .

It will also act towards the right .

Total force created due to contraction in the first spring and extension in the right spring = 8.05 +11.04 N

= 19.09 N .

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Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

vagabundo [1.1K]

Answer:

R_total = 14.57 Ω , V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance

$R_{eq}$ = ∑ $R_{i}$

* For resistance in parallel

1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

R_B = Rb + R4

R_B = 2 + 18

R_B = 20 Ω

Branch C

R_C5 = Rc + R5

R_C5 = 3 + 12

R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

1 / R_BC = 1 / R_B + 1 / R_C5

1 / R_BC = 1/20 + 1/15 = 0.116666

R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

R_total = R_A + R_BC

R_total = 6 + 8.57

R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

10 = V_a + V_BC

10 = i R_a + i R_BC = i (R_a + R_BC)

i = 10 / (R_a + R_BC)

i = 10 / (14.57)

i = 0.6863 A

The current in the series circuit is constant

V_BC = i R_BC

V_BC = 0.6863 8.57

V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

Branch C

V_BC = i R_C5

i = V_BC / R_C5

i = 5.8819 / 15

i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

V_C = i R_C

V_C = 0.39213 3

V_C = 1.176 V

3 0

3 years ago

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$

On multiplying both sides by "100", we get

⇒ $\frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}$

⇒ $\frac{\Delta T}{T}\times 100=0.005787$ (%)

∵ $T=2\pi\sqrt{\frac{l}{g} }$

On substituting the values, we get

⇒ $\frac{\Delta T}{T}$ % = $\frac{1}{2}\times \frac{\Delta l}{l}$ %

On applying cross multiplication, we get

⇒ $\frac{\Delta l}{l}$ % = $2\times \frac{\Delta T}{T}$ %

⇒ = $2\times 0.05787$

⇒ = $0.011574$

⇒ = $0.012$ %

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3 years ago

Help asapp!!!!!!!!!!!!!!

IRINA_888 [86]

Sorry don’t know this one

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3 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

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3 years ago

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kolbaska11 [484]

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2 years ago