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Kisachek [45]
3 years ago
5

On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating

? Ignore any complications, such as loss of water by evaporation. (Enter the magnitude.)
Physics
1 answer:
vichka [17]3 years ago
3 0

Answer:

326149.2 KJ

Explanation:

The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:

Q = m*cv*ΔT

Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.

A 65000 L swimming pool will have a mass of:

65000L *\frac{1m^3}{1000L} * \frac{1000kg}{1m^3} = 65000 kg

The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.

We replace the data and get:

Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ

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A 2 kg block slides on a rough horizontal surface with muk=0.6. It has an initial velocity of 5 m/s. Use g = 10 m/s2
Irina18 [472]

Answer:

360000

Explanation:

4 0
3 years ago
In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
2 years ago
Will mark as brainliest if correct!!!!!!!!!!!!!!!!!!!!!!!!!
Digiron [165]

Answer:

Incident ray

Explanation:

"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."

https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03

3 0
3 years ago
Read 2 more answers
Calculate the mass of -1.5C of electrons​
algol13
The electron is a type of low-mass, very negatively charged with a particle. As such, it can easily be deflected by passing close to other electrons or the positive nucleus of an atom. m = mass of an electron in kg = 9.10938356 × 10-31 kilograms. e = magnitude of the charge of an electron in coulombs = 1.602 x 10-19 coulombs. Hope this helps!
8 0
3 years ago
The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf
kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy  

V0x = V0 cos(θ) 

V0y = V0 sin(θ) 

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2 

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant  

Vx = Vocos(θ) 

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t 

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t 

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2 

3 0
3 years ago
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