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pishuonlain [190]
2 years ago
7

A 17 kg box sitting on a shelf has a potential energy of 350 J. How high is the shelf? Round your answer to the nearest whole nu

mber. The shelf is m high.
Physics
2 answers:
Mars2501 [29]2 years ago
8 0

Answer:

2

Explanation:

got it right on e d g e n u i t y 2020

Free_Kalibri [48]2 years ago
3 0
Potential energy is mass * gravity * height. (m*g*h).

350 = 17*9.8*h   <--350 is its energy, 17kg is its mass, and 9.8 is gravity's acceleration on the object. We now just need to solve for h.

h = 350/(17 * 9.8) =  2.1 meters, which, when rounded to the nearest whole meter, is 2 meters.

The shelf is 2 meters high.

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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

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Peter is a middle-aged, overweight man who is trying to decrease his body circumference. How could the serving size on a nutriti
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3 0
2 years ago
A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t
Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

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Magnitude of Earth's field, B=10^{-4}\ T

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}

\omega=159.15\ rad/s

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

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2 years ago
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4 0
3 years ago
Read 2 more answers
Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
Nonamiya [84]

Answer:

1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0
2 years ago
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