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3 years ago

A 10 kg body is suspended by a rope is pulled

Physics

1 answer:

Margarita [4]3 years ago

5 0

Answer:

The correct option is;

3) 10·√3 kgwt

Explanation:

The parameters given are;

Mass of body, m = 10 kg

Direction of applied force = Horizontal

Angle of inclination of the rope with the vertical, θ = 60°

The component of the tension in the rope equivalent to the force is given by the relation;

T·sin(θ) = F

The component of the rope tension still under gravitational pull is given by the relation;

T·cos(θ) = m·g

Therefore, we have;

$\dfrac{T \cdot sin(\theta)}{T \cdot cos(\theta)} = \dfrac{F}{m \cdot g}$

Which gives;

$\dfrac{sin(\theta)}{cos(\theta)} = tan(\theta) = \dfrac{F}{m \cdot g}$

Therefore;

F = tan(θ) × m×g

Where:

g = Acceleration due to gravity

tan(θ) = √3

m = 10 kg

∴ F = 10·√3 × g = 10·√3 kg-wt.

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What was the direction of the ball’s velocity

Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor? </em>

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$

8 0

3 years ago

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

$\Delta H_{1}=nC_{p}\times\Delta T$

Put the value into the formula

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

$\Delta H_{1}=nC_{v}\times\Delta T$

Put the value into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Hence, The heat flows into the gas during this two-step process is 120 cal.

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4 years ago

If the core of a supernova explosion contains three or more solar masses of matter, the core will most likely become a _____.

Mnenie [13.5K]

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3 years ago

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3. If a 2.0-kilogram mass is raised 0.050 meter

statuscvo [17]

The work done on the mass is approximately 1J

<h3>How to calculate work done on mass</h3>

From the question, we are to determine the work done on the mass

The work done can be calculated from the formula for Potential energy

Work done = P.E = mgh

Where m is the mass

g is the acceleration due to gravity (g = 10 m/s²)

h is the height

From the question,

m = 2.0 kg

h = 0.050 m

Putting the values into the question, we get

Work done = 2.0 × 10 × 0.050

Work done = 1 J

Hence, the work done on the mass is approximately 1J

Learn more on how to calculate work done here: brainly.com/question/14460830

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2 years ago

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Tamiku [17]

Evaporation.............

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3 years ago

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