All categories

2 years ago

An electric heater draws a steady current = 20.0 A on a 120-V line. (a) Calculate how much power does it require.

Physics

1 answer:

babymother [125]2 years ago

8 0

Answer:

The heater power required is 2400 W. The power in the heater can be calculated as the product of the voltage line and the steady current:

$P=V.I$

$P=120 V * 20 A = 2400 VA = 2400 W$

Explanation:

You might be interested in

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago

1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0

2 years ago

An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees

vitfil [10]

Answer:

0.002925 m

Explanation:

Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature

<h2>for aluminium</h2>

α=25×10^-6

Lt = 5(1+25×10^-6×(70-20))

Lt = 5 (1+25×10^-6×50)

Lt = 5 ( 1+0.00125)

Lt = 5×1.00125

Lt =5.00625 m

<h2>for nickel </h2>

α=13.3×10^-6

Lt =5(1+13.3×10^-6×50)

Lt = 5(1+0.000665)

Lt =5.003325 m

hence difference in length =5.00625-5.003325

= 0.002925 m

3 0

2 years ago

A 17 g audio compact disk has a diameter of 12 cm. The disk spins under a laser that reads encoded data. The first track to be r

sladkih [1.3K]

Answer:

0.00066518 Nm

Explanation:

v = Velocity = 1.2 m/s

r = Distance to head = 2.3 cm

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 2.4 s

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{1.2}{0.023}\\\Rightarrow \omega=52.17391\ rad/s$

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{52.17391-0}{2.4}\\\Rightarrow \alpha=21.73912\ rad/s^2$

Torque

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mR^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}0.017\times 0.06^2\times 21.73912\\\Rightarrow \tau=0.00066518\ Nm$

The torque of the motor is 0.00066518 Nm

6 0

2 years ago