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maria [59]
2 years ago
5

Which statement is true? The speed of sound in air is inversely proportional to the temperature of the air. The speed of sound i

n air is directly proportional to the temperature of the air. The speed of sound in air is twice the temperature of the air. The speed of sound in air is not affected by the temperature of the air.
Physics
2 answers:
Ganezh [65]2 years ago
7 0

Answer:

the speed of sound in air is directly proportional to the temperature of the air

Explanation:

sasho [114]2 years ago
6 0
<span> the speed of sound in air is directly proportional to the temperature of the air. The speed of sound depends on the temperature of the surrounding air, this can be represented by a speed of sound in air formula: v = 331m/s + 0.6m/s/C * T (where T is temperature)</span>
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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius r = 0.330

m

Force F = 290 N

Angular acceleration \alpha  = 0.814 \frac{rad}{s^{2} }

From the formula of torque,

 Γ = I\alpha                                        (1)

 Γ = rF                                       (2)

rF = I \alpha

Find momentum of inertia I from above equation,

I = \frac{rF}{\alpha }

I = \frac{0.330 \times 290}{0.814}

I = 117.56 Kg. m^{2}

Find the momentum inertia of disk,

 I = \frac{1}{2}  Mr^{2}

M = \frac{2I}{r^{2} }

M = \frac{2 \times 117.56}{(0.330)^{2} }

M = 2159.045 Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0
3 years ago
A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
3 years ago
A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt
Yuri [45]
The total power emitted by an object via radiation is:
P=A\epsilon \sigma T^4
where:
A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW
So, the correct answer is D.
6 0
3 years ago
While driving on a rural road, your right wheels run off the pavement. You should hold the steering wheel firmly and
Vsevolod [243]

Answer:

The answer is C. Steer in a straight line while gently slowing down

Explanation:

The following are advised when your cars go off the pavement while driving;

firstly, Do not panic.

ensure you hold on to your steering wheel tightly.

keep Steering straight ahead.

ensure you Stay on the shoulder.

Ease up on the accelerator and brake gently.

When you  know you can safely do so, turn back on the road at a much lower speed.

3 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
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