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notka56 [123]
4 years ago
12

Explain the difference between mass and weight and how they are measured.

Physics
2 answers:
ASHA 777 [7]4 years ago
4 0
There is a basic difference<span>, because </span>mass<span> is the actual amount of material contained in a body and is </span>measured<span>in kg, gm, etc. Whereas </span>weight<span> is the force exerted by the gravity on that object mg. Note that </span>mass<span> is independent of everything but </span>weight<span> is different on the earth</span>
Lunna [17]4 years ago
3 0
Mass is a coherent, typically large body of matter with no definite shape. Weight is a body's relative mass or the quantity of matter contained by it, giving rise to a downward force; the heaviness of a person or thing. <span>You can </span>measure mass<span> using a balance. Weight is measured with a scale.
</span>
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What happen to the bulb when it is in series connection?
coldgirl [10]
If the bulb is in series with something else, then . . .

--  The brightness of the bulb depends on the <em>other</em> device in the circuit. 

--  If the other device is designed to use <em>less power</em> than the bulb, then the
other device gets <em>more power</em> than the bulb gets.

--  If the other device is designed to use <em>more power </em>than the bulb, then the
other device gets <em>less power</em> than the bulb gets.

--  If the other device is removed from the circuit, then the bulb doesn't light at all.

This description of the often-screwy behavior of a series circuit may partly explain
why the electric service in your home is not a series circuit.




8 0
3 years ago
Read 2 more answers
The moon is slowly moving away from the earth. What will eventually happen to the gravitational pull force between the earth and
sesenic [268]
The gravitational pull will reduce, because the moon is moving further away from the centre of the earth. In other words, the further an object is from the centre of the earth, the weaker the gravitational pull that will be exerted on it. 
6 0
3 years ago
Please help with Physics Circuits!
Zigmanuir [339]
1) Let's start by calculating the equivalent resistance of the three resistors in parallel, R_2, R_3, R_4:
\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}
From which we find
R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega
So, the correct answer is D) 


2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega

And these are connected in series with a resistor of 10 \Omega, so the equivalent resistance of the circuit is
R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega

And by using Ohm's law we find the current in the circuit:
I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A
So, the correct answer is C).


3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega

And by using Ohm's law we find the current flowing in the circuit:
I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V
So, the correct answer is D).


4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}
From which we find
R_{23} = 7.55 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is:
R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega

The current in the circuit is given by Ohm's law
I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A

Now we can compare the voltage drops across the resistors. Resistor 1:
V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V
Resistor 4:
V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V

So, the greatest voltage drop is on resistor 1, so the correct answer is D).


5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
P_3 = I_3^2 R_3 =  \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.
5 0
3 years ago
Read 2 more answers
Generators are device that convert_____energy into____energy
KiRa [710]

Answer:

A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy.

Explanation:

5 0
4 years ago
A wave is traveling in a string at 58 m/s. when the tension is then increased 17%, what will be the resulting wave speed?
Zepler [3.9K]

the resulting wave speed=62.7 m/s

Explanation:

The speed of a wave is given by

v=\sqrt{\frac{T}{\mu} }

T= tension, v= velocity= 58 m/s

when the tension is increased by 17%, new tension= 1.17 T

V'=\sqrt{\frac{1.17T}{\mu}}

V'=1.082 \sqrt{\frac{T}{\mu}}

V'=1.082 V

V'=1.082 (58)

V'=62.7 m/s

7 0
3 years ago
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