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ivolga24 [154]
3 years ago
5

If this experiment is repeated at a place3 meters above the sea level the acceleration due to gravity is expected to

Physics
1 answer:
weqwewe [10]3 years ago
8 0

The question doesn't describe any experiment.  If the same experiment is repeated, no matter how many times, the acceleration due to gravity will remain the same as it was during the non-existent original experiment, and will have no effect on anything.

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How much time will it take a car traveling at 88 km/hr (55 mi/hr) to travel 500 km?
Alexeev081 [22]
The answer is5.68 hours
8 0
3 years ago
A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrof
Vinil7 [7]

The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

<h3>What is the specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol  m₁ = 200-gram

The temperature of alcohol, T₁ =  -6°C.

Mass of liquid sample of water  m₂ = 400-gram

The temperature of the water, T₂=  20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

<h3 />

\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72  \ kJ /kg ^0 C

Hence the specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.

#SPJ1

6 0
2 years ago
A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an
Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

\omega = 2\pi f

\omega = 2\pi (2.5)

\omega = 5\pi rad/s

The angular displacement is given as the form:

\theta (t) = \theta_0 cos(\omega t)

In the equlibrium we have to t=0, \theta(t) = \theta_0 and in the given position we have to

\theta(t) = \theta_0 cos(5\pi t)

Derived the expression we will have the equivalent to angular velocity

\frac{d\theta}{dt} = 2.7rad/s

Replacing,

\theta_0(sin(5\pi t))5\pi = 2.7

Finally

\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°

Therefore the maximum angular displacement is 9.848°

6 0
3 years ago
A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,
ZanzabumX [31]

Answer:

5) 13 revolutions (approximately)

Explanation:

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2α*θ Formula (1)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

Data:

ω₀ = 18 rad/s

ωf = 0

α = -2 rad/s²  ; (-) indicates that the wheel is slowing

Revolutions calculation that turns the wheel until it stops

We apply the formula (1)

ωf²= ω₀² + 2α*θ

0 = (18)² + 2( -2)*θ

4*θ =  (18)²

θ =  (18)²/4 = 81 rad

1 revolution = 2π rad

θ = 81 rad * 1 revolution / 2πrad

θ =  13 revolutions approximately

8 0
3 years ago
The normal is a line perpendicular to the reflecting surface at the point of incidence.
ladessa [460]

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.

4 0
3 years ago
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