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2 years ago

8

What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0. 5 A in the bulb? 0. 0042

0. 5 60 240.

1 answer:

OlgaM077 [116]2 years ago

4 0

Answer is 240 ohms

Use the equation V=IR, rearranging it to make R (resistance) the subject.

R = V/I

Substitute the values:
V = 120
I = 0.5

Resistance = 120 / 0.5

Resistance = 240 ohms

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You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta

Romashka-Z-Leto [24]

Answer:

a) 3.66 s

b) 124.4 m

c) 3.12s

Explanation:

Given that

Speed of the Red Car, v₁ = 34 m/s

Speed of the Blue Car, v₂ = 28 m/s

Distance between the two cars, d = 22 m

The difference between the speed of the cars is: 34 - 28 = 6 m/s

From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.

1)

If we divide the distance by the difference in speed. This becomes

d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.

2

From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.

v = d * t = 34 * 3.66 = 124.4

Therefore the red car travels at 124.4 m before catching up to the blue car.

3

If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.

S = ut + ½gt², where

S = 22

u = 6

t = ?

g = 2/3

22 = 6t + 1/3t²

By using the quadratic formula, we find out the two answers listed below

t1 = 3.12 s

t2 = - 21.12 s

We all know that negative time is not possible, so the answer is t1. At 3.12 seconds

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3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

$kA^2 =mV^2 +ky^2$

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

$1700*A^2=5.3*1.7^2 +1700*(0.045)^2$

Solving for A,

$A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}$

$A ^2 = 0.011035$

$A=0.105 m \approx 10.5 cm$

PART C) Finally, the total mechanical energy is given by the equation

$E = \frac{1}{2}kA^2$

$E=\frac{1}{2}1700*(0.105)^2$

$E= 9.3712 J$

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3 years ago

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Monica [59]

The magnitude of the action–reaction pair between the two boxes (A and B) will be "18.2 N".

According to the question,

Mass of box A,

$m_A = 9\ kg$

Mass of box B,

$m_B = 2 \ kg$

Horizontal force,

$F_{app} = 100 \ N$

From the Newton's law,

→ $F_{app} = (\frac{F_{app}}{m_A+m_B} )a$

or,

→ $a = \frac{F_{app}}{(m_A+m_B)}$

Bu substituting the values, we get

→ $= \frac{100}{9+2}$

→ $= \frac{100}{11}$

→ $9.10 \ m/s^2$

We can see that between the two boxes, the action-reaction pair exist.

then,

→ $F_{action-reaction} = m_b \ a$

→ $=2\times 9.10$

→ $= 18.2 \ N$ (magnitude)

Thus the above solution is appropriate.

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hram777 [196]

Armature is the correct answer.

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Because it has both magnitude and direction. When an object has positive acceleration, the acceleration occurs in the same direction as the movement of the object.

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