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a_sh-v [17]
2 years ago
13

What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?

Physics
1 answer:
madam [21]2 years ago
4 0

Answer:

there are 25 kg objective travelling at 2m/s to the right.

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PLEASEEE HELP, thank you :)
telo118 [61]

Answer:

The answer is B.

Explanation:

Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :

Total resistance = 20Ω + 40Ω + 60Ω

= 120Ω

Next, we have to find out that 1Ω is equal to how many voltage by dividing :

120Ω = 60V

1Ω = 60V ÷ 120

1Ω = 0.5V

Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :

1Ω = 0.5V

20Ω = 0.5V × 20

20Ω = 10V

8 0
3 years ago
8. When you see a stopped local bus (1 point)
elena55 [62]

Answer: be alert for pedestrians near the bus.

Explanation: Due to road accidents many Governments around the world has adopted and put in place certain rules and regulations with regards to road safety, this is so to prevent the or reduce the chances of accidents happening.

Road safety rules are rules and guidelines put in place by Government in order to prevent road accidents and maintain a free flow of traffic. An example of such rules is 'be alert for pedestrians near the bus ' when approaching a local bus that is stopped.

5 0
3 years ago
A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
A Nichrome wire 44 cm long and 0.30 mm in diameter is connected to a 3.1 V flashlight battery. What is the electric field inside
Alexeev081 [22]

Answer:

7.05 Volts/m

Explanation:

L = length of the Nichrome wire = 44 cm = 0.44 m

V = Potential difference across the end of the wire = battery voltage = 3.1 Volts

E = magnitude of electric field inside the wire

Magnitude of electric field inside the wire is given as

E = \frac{V}{L}

Inserting the values

E = \frac{3.1}{0.44}

E = 7.05 Volts/m

4 0
3 years ago
Light that has a wavelength of 554 nm is incident on a single slit that is 10 μm wide. Determine the angular location of the fir
VMariaS [17]
66.3 = 3% 9.528 correct 3.176 correct
7 0
4 years ago
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