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inn [45]
3 years ago
9

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

the volume is constant, and it is heated to a temperature of 378 K. Find the net force on each wall of the box.
Physics
1 answer:
const2013 [10]3 years ago
5 0

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}

we know that

force = pressure * area

therefore force is

F =(\frac{T_{1}}{T_{2}}*P_{1})A

F =(\frac{378}{278}*101325)(17*10^{-2})^{2}

F = 3.98 kN

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a van moves with a constant speed of 79 km/h how long will it take to travel a distance of 502 kilometers
Mazyrski [523]

Answer:

6.35hours

Explanation:

s=vt

t=s/v=502/79=6.35hours

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Svetllana [295]

D: a force equal to the force

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A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
Read 2 more answers
To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If
arlik [135]

Answer:

Part a)

a = 1260.3 m/s^2

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

v_f^2 - v_i^2 = 2 a d

v_1^2 - 0^2 = 2(9.81)(4.0)

v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

v_f^2 - v_i^2 = 2 a d

0 - v_2^2 = 2(-9.81)(2.00)

v_2 = 6.26 m/s

Part a)

Average acceleration is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}

a = 1260.35 m/s^2

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0
3 years ago
Explain the universal law of gravity and it's calculate from as well as the value of g​
horrorfan [7]

Answer:

According to the Newton's law of gravitational every object in the universe attracts every other objects with a force which is called gravitational force.This gravitational force is (i) directly proportional to the product of their masses and (ii) inversely proportional to the square of the distance between their centres.

Explanation:

Newton's law of gravitational is called the universal law because it is applicable to all the bodies either terrestrial or celestial having any shape,size,mass or at any distance apart with any medium between them,at any time(past,present or future).

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