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2 years ago

Seema knows the mass of basketball. What other information is needed to find the balls potential energy

Physics

2 answers:

Lelu [443]2 years ago

8 0

Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field. In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy $U$ will be:

$U=mgh$

Where:

$m$ is the mass of the ball

$g=9.8 m/s^{2}$ is the acceleration due gravity (assuming the ball is on the Earth surface)

$h$ is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

Zarrin [17]2 years ago

6 0

Answer:

C. acceleration due to gravity and the height the ball reaches

Explanation:

because the potential energy equation is PE=mgh so, if we have m (mass) then all we need is g (gravity) and h (height) so your answer is c

hope this helps☺️

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A solid plate, with a thickness of 15 cm and a thermal conductivity of 80 W/m·K, is being cooled at the upper surface by air. Th

Nezavi [6.7K]

To solve this problem we will use the two principles that are visible according to the phenomena described in the problem: Heat transfer by conductivity and Heat transfer by convection.

This thermal transfer will be equivalent and with it we can find the value asked.

<em>Note: We will assume that the temperature value at the plate surface is: 60 ° C (For the given value of 650 50)</em>

For Thermal Transfer by Conduction

$Q_{cn} = -kA \frac{\Delta T}{\Delta x}$

$Q_{cn} = -kA \frac{T_1-T_2}{L}$

Where,

k = Thermal conductivity

A = Cross-sectional Area

$T_2$ = Temperature of the bottom surface

$T_1$ = Temperature of the top surface

L = Length

Replacing we have that

$Q_{cn} = -(80W\cdot K)(A)\frac{50\°C-60\°C}{15cm\frac{1m}{100cm}}$

$Q_{cn} = 5333.33A$

For Thermal Transfer by Convection

$Q_{cv} = hA(T_1-T_{\infty})$

Where,

h = Convection heat transfer coefficient

$T_{\infty}$ = Surrounding temperature

A = Surface Area

Replacing we have that

$Q_{cv} = hA(50\°C-10\°C)$

$Q_{cv} = 40hA$

Since the rate of heat transfer by convection is equal to that given by conduction we have to:

$Q_{cn}=Q_{cv}$

$5333.33A = 40hA$

$h = 133.33W/m^2\cdot K$

It is stated that the typical values of forced convection of gases lies in the range of $(25-250)W/m^2\cdot K$ . The obtained value is reasonable for forced convection of air.

7 0

2 years ago

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider dives to a height of 278 mete

Verdich [7]

Answer:

<em>The glider's new speed is 68.90 m/s</em>

Explanation:

<u>Principle Of Conservation Of Mechanical Energy</u>

The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

$\displaystyle K=\frac{1}{2}mv^2$

The gravitational potential energy when it's at a height h from the zero reference is

$U=mgh$

The total mechanical energy is

$M=K+U$

$\displaystyle M=\frac{1}{2}mv^2+mgh$

The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.

The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

$\displaystyle M_1=\frac{1}{2}m(45.2)v_o^2+m(9.8)(416)$

Operating in terms of m

$\displaystyle M_1=1021.52m+4076.8m$

$\displaystyle M_1=5098.32m$

Then we know the glider dives to 278 meters and we need to know their final speed, let's call it $v_f$ . The final mechanical energy is

$\displaystyle M_2=\frac{1}{2}mv_f^2+m(9.8)(278)$

Operating and factoring

$\displaystyle M_2=m(\frac{1}{2}v_f^2+2724.4)$

Both mechanical energies must be the same, so

$\displaystyle m(\frac{1}{2}v_f^2+2724.4)=5098.32m$

Simplifying by m and rearranging

$\displaystyle \frac{v_f^2}{2}=5098.32-2724.4$

Computing

$v_f=\sqrt{4747.84}=68.90\ m/s$

The glider's new speed is 68.90 m/s

8 0

2 years ago

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

$v = - v_o \sqrt{70 \%}$ → <em>negative sign due to ball rebounds</em>

$v = - v_o \sqrt{0.7} \texttt{ m/s}$

$\texttt{ }$

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Average Speed of Plane : brainly.com/question/12826372
Impulse : brainly.com/question/12855855
Gravity : brainly.com/question/1724648

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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2 years ago

Karissa is conducting an experiment on the amount of salt that dissolves in water at different temperatures. She repeats her tes

Oksanka [162]

Answer:

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2 years ago