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Zarrin [17]
3 years ago
11

The earth's hydrosphere includes which of the following?

Physics
2 answers:
Andre45 [30]3 years ago
6 0
I had this question too lol the answer is D. aka All Of Them!
Oduvanchick [21]3 years ago
5 0
The answer is all of these
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Give an example of a positively charged object
allochka39001 [22]
When a charged balloon is brought near a conducting can, the electrons are pushed to the far side of the can. If another object touches the side of the can opposite the balloon, electrons flow into the other object, leaving the can positively charged
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2 years ago
A conducting wire formed in the shape of a right triangle with base b = 25 cm and height h = 66 cm and having resistance R = 1.1
NemiM [27]

Answer:

\omega=4.19 \,rad.s^{-1}

Explanation:

To find:

angular frequency of rotation, ω=?

Given that:

  • base of wire triangle, b = 25 cm
  • height of wire triangle, h = 66 cm
  • electrical resistance , R = 1.1 Ω
  • time taken to make 1 rotation about the height through the base,T= 1.5s
  • magnetic field in +z direction out of the screen, B = 1.8 T

Since the wire does not carry any current therefore the magnetic field will have no impact on it.

Now, as we know:

angular frequency

\omega= \frac{2\pi}{T}

\omega=4.19 \,rad.s^{-1}

5 0
3 years ago
A student slides a book across a desk, with a velocity of +8 m/s. When her friend catches the book, it has a velocity of +7.4 m/
777dan777 [17]

kinematic equation

v^2=u^2+2*a*s

7.4^2=8^2-2*5.6^2*width

(7.4^2-64)/(2*5.6^2)=-width


3 0
3 years ago
I need help on questions 4-7 plz
Drupady [299]
Sorry don't know. But let me try to figure out
7 0
4 years ago
Please help with Physics Circuits!
Vaselesa [24]
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.
8 0
3 years ago
Read 2 more answers
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