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sukhopar [10]
3 years ago
12

Please help please help

Physics
1 answer:
Inessa [10]3 years ago
6 0

Answer: p= m/v so 90kg/.075m^3 = 1,200

2a. .35 m 1.1 m and .015 m

2b. 35 cm x 110 cm x 1.5 cm = 5,775 cm^3 = 57.75 m^3

mass= pv

2700•57.75= 155,925 kg

mass= 155,925 kg

volume= 57.75 m^3

Explanation: physics

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An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res
murzikaleks [220]

Answer:

(a) V=11.86\ V

(b) V=9.76\ V

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

V=R.I

(a) The electromagnetic force of the battery is \varepsilon =12.6\ V and its internal resistance is R_i=0.06\ \Omega. Knowing the equivalent resistance of the headlights is R_e=5.2\ \Omega, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)

Solving for i

\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A

i=2.28\ A

The potential difference across the headlight  bulbs is

V=\varepsilon  -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V

V=11.86\ V

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

V=\varepsilon  -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V

5 0
3 years ago
A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur
Anna35 [415]

Answer:

\dot{W_{H} } = 4244.48 Btu/h

Explanation:

Temperature of the house, T_{H} = 70^{0} F

Convert to rankine, T_{H} = 70^{0}+ 460 = 530 R

Heat is extracted at 40°F i.e T_{L} = 40^{0}F  = 40 + 460 = 500 R

Calculate the coefficient of performance of the heat pump, COP

COP = \frac{T_{H} }{T_{H} - T_{L}  } \\COP = \frac{530 }{530 - 500  }\\ COP = \frac{530}{30} \\COP = 17.67

The minimum power required to run the heat pump is given by the formula:

\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\...............(*)

Where the heat losses from the house, \dot{Q_{H} } = 75,000 Btu/h

Substituting these values into * above

\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h

3 0
3 years ago
What are at least three types of energy involving a microwave
timama [110]
I'm pretty sure what you are trying to ask for is radiative energy, light energy, and electronic energy.
Radiative since the microwave is releasing radiation,
Light since there is light inside the microwave,
Electronic since it is plugged in and uses electricity.
You can also use sound, but I don't think every microwave makes sound. 
8 0
3 years ago
Which of these is not a type of force?
coldgirl [10]
Inertia
the awnswer is inertia b
6 0
3 years ago
Read 2 more answers
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
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