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skelet666 [1.2K]
1 year ago
14

The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

builds a pendulum by attaching one end of a string of known length to the ceiling and the other end to a small solid sphere. The sphere is pulled back so the string makes a 5 angle with the vertical and released from rest with a theoretical amplitude of Oscillation At The student measures the resulting experimental period Tg and after the sphere makes several oscillations, measures the experiment oscillation. In the absence of air resistance, T T and A. Which of the following best explains how air resistance affects the results of the experiment?
A) with resistance because the sphere will speed up
B) With air resistance As , because although the sphere slows down, it also will travel for a greater time
C) With resistance because the sphere will travel a greater distance
D) With air resistance, 7>7y, because the sphere will slow down
E) with resistance because although the sphere slows down, also will travel shorter distance
Physics
1 answer:
astra-53 [7]1 year ago
4 0

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

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An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The
Zolol [24]

Answer:

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

4 0
2 years ago
A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove
True [87]

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can m_1=4 kg

mass of second can m_2=3 kg

let T be the Tension in the rope

For  4 kg can

4g-T=4a

T=4(g-a)----1

For 3 kg can

T-3g=3a

T=3(g+a)----2

From 1 and 2

4(g-a)=3(g+a)

g=7a

a=\frac{g}{7}

So time taken to cover 18 m is

h=ut+\frac{at^2}{2}

18=0+\frac{g\cdot t^2}{7\times 2}

t^2=\frac{18\times 2\times 7}{g}

t=5.07 s

5 0
2 years ago
What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

3 0
2 years ago
When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe
ch4aika [34]

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0
2 years ago
A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

4 0
2 years ago
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