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Nataly [62]
3 years ago
7

A cannonball with a mass of 8.3 kg is fired from a cannon that has a mass of 949.3 kg. The cannonball is fired with a speed of 1

87.9 m/s, and it takes 0.0 seconds for the cannonball to leave the cannon.
what is the recoil speed?
Physics
1 answer:
Fittoniya [83]3 years ago
6 0
<span>conservation of momentum
mass(ball) × speed(ball) = mass(cannon) × speed(cannon)
(8.3 kg) × (187.9 m/s) = (949.3 kg) × v
1.64 m/s = v
</span>
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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
what would happen if a paper clip that is not a permanent magnet was brought close to the south pole of a permanent magnet?
Diano4ka-milaya [45]

Answer:

Because this pole is different from the original pole that magnetized the paperclips, the top paperclip will become demagnetized in the opposite direction and the bottom paperclip will be repelled away

8 0
3 years ago
Let’s say you have a cart of some mass and when pushed with 10N of force, the cart accelerates at 5.0 m/s/s. If you were to push
Korolek [52]

Answer:

a. The acceleration would increase.

Explanation:

because we know that

F=ma

m= mass and a= acceleration

so Mass is same for the cart in any situation that's why only acceleration could increase.

7 0
3 years ago
Read 2 more answers
8 ( 3b + 3 ) = 5b + 5
Sophie [7]
You first distribute 8 through the parenthesis so it’s 24b+24=5b+5
Then collect the like terms to calculate 24b-5b=5-24 you then subtract so now it’s 19b=-9
So now you divide and your answer is b=-1
4 0
3 years ago
An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If th
stepan [7]

Answer:

Number of photons travel through pin hole=6.4*10^{17}

Explanation:

First we will calculate the energy of single photon using below formula:

E=\frac{h*c}{λ}

Where :

h is plank's constant with value 6.626*10^{-34} J.s

c is the speed of light whch is3*10^{8}

λ is the wave length = 532nm

E=\frac{6.6268*10^{-34}* 3*10^{8}}{532nm}

E=3.73*10^{-19}J

Number of photons emitted per second:

\frac{5J/s}{1 photon/3.73*10^{-19} }

Number of photons emitted per second=1.34*10^{19} photons/s

\frac{A-hole}{A-beam}=\frac{\frac{pie*d-hole^2}{4}}{\frac{pie*d-beam^2}{4} }

Where:

A-hole is area of hole

A-beam is area of beam

d-hole is diameter of hole

d-beam is diameter if beam

\frac{A-hole}{A-beam}=\frac{d-hole^2}{d-beam^2}

\frac{Ahole}{A-beam}=\frac{1.22^2}{5.5^2}

\frac{A-hole}{A-beam}=\frac{144}{3025}

Number of photons travel through pin hole=1.34*10^{19} *\frac{144}{3025}

Number of photons travel through pin hole=6.4*10^{17}

7 0
3 years ago
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