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snow_tiger [21]
3 years ago
12

If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t

o the piano before it reaches the bottom of the ramp?

Physics
1 answer:
solong [7]3 years ago
3 0
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s

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An electric heater carries a current of 13.5 A when operating at a voltage of 120V . What is the resistance of the heater?
Aloiza [94]

An electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have  the resistance of the heater as 8.89 ohm.

<h3>What is Ohm's law?</h3>

Ohm's law  can be described as the law in physics that   focus on how voltage or potential difference between two points is directly proportional to the current or electricity that is moving in the direction of the resistance.

An this is also  directly proportional to the resistance of the circuit, hence it can be calculated using the Ohm's law is V=IR.

from V=IR

R=V/I

= 120/13.55

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Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu
Rus_ich [418]

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is I  = 3.535 *10^{-6} \  W/m^2

Explanation:

From the question we are told that

    The temperature is  T = 2.81 \ K

Now  According to Stefan's law

        Power(P) =  \sigma  *  A  * T^4

Where  \sigma is the Stefan Boltzmann constant with value  \sigma  =  5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        I  =  \frac{P}{A}

=>      I  =  \frac{\sigma *  A  * T^4}{A}

=>      I  =  \sigma  *  T^4

substituting values

      I  = 5.67 *10^{-8}  *  (2.81)^4

       I  = 3.535 *10^{-6} \  W/m^2

       

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3 years ago
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