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AleksAgata [21]
3 years ago
12

I'm literally so dumb and about to fail my ap physics exam please help

Physics
1 answer:
Goshia [24]3 years ago
5 0
Ok bro, so what’s the question lol
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How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t
tekilochka [14]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

Q=mC\Delta T\\\\Q=\rho VC\Delta T

(m=\rho V)

where,

Q = amount of heat required = ?

m = mass

\rho = density of air = 1.20kg/m^3

V = volume of air

C = specific heat of air = 1006J/kg^oC

\Delta T = change in temperature = 10.0^oC

Now put all the given values in above formula, we get:

Q=\rho VC\Delta T

Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)

Q=1.449\times 10^6J

Now we have to calculate the time required.

Formula used :

t=\frac{Q}{P}

where,

t = time required = ?

Q = amount of heat required = 1.449\times 10^6J

P = power = 1500 W

Now put all the given values in above formula, we get:

t=\frac{1.449\times 10^6J}{1500W}

t=966s\times \frac{1min}{60s}=16.1min

Thus, the time required is, 16.1 minutes.

5 0
3 years ago
A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

M_{before} = M_{after}

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 =  velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

3 0
3 years ago
Sphere A has a charge of +2 x 10 to the power of -6 coulomb and is brought into contact with a similar sphere, B which has a cha
Setler [38]
When sphere A and B are brought in contact and separated, charge on each sphere becomes [2x10^-6 + (-4x10^-6)]/ 2 = -1x10^-6 C.

That is, charge is equally separated and is the average of charges on both spheres. The reason behind equal charge on both spheres after separation is, when they are kept in contact, their potential difference becomes same.
7 0
3 years ago
A box is against a wall. A person pushes on the box, but the box does not move. Is this
Serhud [2]

Answer:

it is a force

Explanation:

a force is a push or a pull

7 0
2 years ago
Read 2 more answers
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
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