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Reptile [31]
2 years ago
10

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

masses are gently released, what is the resulting acceleration of the masses
Physics
1 answer:
AleksAgata [21]2 years ago
5 0

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

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The average kinetic energy of the molecules of an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsiu
Westkost [7]

Answer:

A) T1 = 566 k = 293°C

B) T2 = 1132 k = 859°C

Explanation:

A)

The average kinetic energy of the molecules of an ideal gas is givwn by the formula:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy

K = Boltzman Constant

T = Absolute Temperature

At 10°C:

K.E = K10

T = 10°C + 273 = 283 K

Therefore,

K10 = (3/2)(K)(283)

FOR TWICE VALUE OF K10:

T = T1

Therefore,

2 K10 = (3/2)(K)(T1)

using the value of K10:

2(3/2)(K)(283) = (3/2)(K)(T1)

<u>T1 = 566 k = 293°C</u>

<u></u>

B)

The average kinetic energy of the molecules of an ideal gas is given by the formula:

K.E = (3/2)KT

but K.E is also given by:

K.E = (1/2)(m)(vrms)²

Therefore,

(3/2)KT = (1/2)(m)(vrms)²

vrms = √(3KT/m)

where,

vrms = Root Mean Square Velocity of Molecule

K = Boltzman Constant

T = Absolute Temperature

m = mass

At

T = 10°C + 273 = 283 K

vrms = √[3K(283)/m]

FOR TWICE VALUE OF vrms:

T = T2

Therefore,

2 vrms = √(3KT2/m)

using the value of vrms:

2√[3K(283)/m] = √(3KT2/m)

2√283 = √T2

Squaring on both sides:

(4)(283) = T2

<u>T2 = 1132 k = 859°C</u>

8 0
2 years ago
A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow
leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

7 0
2 years ago
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
11111nata11111 [884]

Answer:

\sigma=0.014\ C/m^2

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2

So, the surface charge density on the sphere is 0.014\ C/m^2.

7 0
2 years ago
Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c
ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0
2 years ago
(a) You wish to determine the height of the smokestack of a local coal burning power plant. You convince a member of the mainten
damaskus [11]

SHİNE LİKE A BRİGHT STAR

❄NEVER GİVE UP❄

☻NEVER SAY NEVER☻

4 0
2 years ago
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