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Vanyuwa [196]
2 years ago
15

A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially

at rest, is dropped onto the car.
What will be the car's new speed?
Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
Alchen [17]2 years ago
5 0

The new speed of car is 10.9 m/s

<h3 />

According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

m_1u_1 + m_2u_2 = (m_1 + m_2) v

v =\frac{m_1u_1}{m_1+m_2}

v = \frac{7950*15}{7950+2950}

v= 10.9 m/s

Therefore, the new speed of care is 10.9 m/s

Learn more about momentum here:

brainly.com/question/22257327

#SPJ1

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So A must have the smallest wavelengths.

4 0
3 years ago
Read 2 more answers
The conduction velocity of an axon is determined by:
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4 0
2 years ago
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan
saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

<u>T₂ = 95.56°C</u>

7 0
3 years ago
Can I PLEASE get some help?
Rus_ich [418]
1.) C
2.) B
3.) D
4.) B
Good luck with your work!
3 0
3 years ago
A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
3 years ago
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