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4 years ago

A marble rolls off a table with a speed of 2.30 m/s. If the table is 1.12 m high, how far from the

Physics

1 answer:

Vlada [557]4 years ago

6 0

Answer:

1.10 m

Explanation:

First, find the time it takes to land.

Given, in the y direction:

Δy = 1.12 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(1.12 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.478 s

Next, find the distance traveled in that time.

Given, in the x direction:

v₀ = 2.30 m/s

a = 0 m/s²

t = 0.478 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.30 m/s) (0.478 s) + ½ (0 m/s²) (0.478 s)²

Δx = 1.10 m

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Does a light bulb at a temperature of 2500K produce as white a light as the sun at 6000K

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Answer:

No. ... UV light causes sunburn, whereas visible light does not.

Explanation:

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3 years ago

A parallel-plate capacitor is connected to a battery. The space between the two plates is empty. If the separation between the c

Maru [420]

Answer:

The final stored energy will become half.

Explanation:

We know that stored energy in the capacitor is given as

$E=\dfrac{1}{2}CV^2$

C=capacitance

V=Voltage difference

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$C=\dfrac{\varepsilon A}{d}$

d=Distance between plates

A=Area

$E=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d}\times V^2$

If the distance between plates get double ,say d' = 2 d

Then stored energy

$E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d'}\times V^2$

$E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{2d}\times V^2$

$E'=\dfrac {E}{2}$

Therefore the final stored energy will become half.

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3 years ago

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 39.0 ∘ abov

zvonat [6]

Answer:

kinetic energy of glider = -0.376 j

Explanation:

Original distance by which the spring is compressed can be determine by using law of conservation of energy (As spring slide from glider, it gain some potential energy. On the top of track there is glider’s potential energy exist)

$\frac{1}{2} kx^2 = mgh$

where x is compressed distance by spring

height of track is calculated as

$sin\theta \frac{h}{l}$

$h = l sin\theta$

where l is distance covered by glider, theta angle is the angle of inclination of track

therefore, we have formula for x as

$x = \sqrt{\frac{2mglsin\theta}{k}}$

$x = \sqrt{\frac{2\times 0.05 \times 9.8 \times 1.50\times sin39}{600}}$

x = 0.039 m

kinetic energy of glider = potential energy by spring - potential energy at 1.50 distance

$=( \frac{1}{2} \times 600\times 0.039^2) - (0.05 \times 9.8 \times 1.5 \times sin 39)$

kinetic energy of glider = -0.376 j

8 0

4 years ago

Four ways to increase magnitude of current in dynamo

8090 [49]

Answer:

hmm

Explanation:

By increasing the number of turns in the coil, strength of magnetic field, speed of rotation of the coil in the magnetic field and by decreasing the distance between the coil and the magnet the magnitude of the induced e.m.f. can be increased in generator/dynamo.

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3 years ago

g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor

Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1, and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0

3 years ago