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kumpel [21]
3 years ago
6

A bus and a car have an inelastic head-on collision. The bus has a mass of 1.5 × 103 kilograms and an initial velocity of 20 met

ers/second. The car has a mass of 9.5 × 102 kilograms and an initial velocity of -26 meters/second. What is their total momentum after the collision?
Physics
2 answers:
makkiz [27]3 years ago
7 0

Answer:

+5300 kg m/s

Explanation:

In any type of collision, the total momentum is conserved. Therefore, we can just calculate the total momentum before the collision, and the final momentum will be equal to the initial one.

The total momentum before the collision is:

p_i = m_1 u_1 + m_2 u_2

where

m_1 = 1.5 \cdot 10^3 kg is the mass of the bus

m_2 = 9.5\cdot 10^2 kg is the mass of the car

u_1 = 20 m/s is the initial velocity of the bus

u_2 = -26 m/s is the initial velocity of the car

Substituting the numbers, we find

p_1 = (1.5 \cdot 10^3 kg)(20 m/s)+(9.5\cdot 10^2 kg)(-26 m/s)=5300 kg m/s

And since the total momentum is conserved, this is also the final momentum after the collision.

Ilya [14]3 years ago
6 0

Answer:

p = 5300 kg-m/s

Explanation:

It is given that,

Mass of the bus, m_1=1.5\times 10^3\ kg

Initial velocity of the bus, u_1=20\ m/s

Mass of the car, m_2=9.5\times 10^2\ kg    

Initial velocity of the car, u_2=-26\ m/s

Let p is their total momentum after the collision. It can be given by :

p=m_1u_1+m_2u_2

p=1.5\times 10^3\times 20+9.5\times 10^2\times (-26)

p = 5300 kg-m/s

So, their total momentum after the collision is 5300 kg-m/s. Hence, this is the required solution.

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lakkis [162]

Answer:

T = 17.26 ^oC

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

Q_1 = Q_2

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so we will have

32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)

so we have

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so we have

16.1 T = 277.87

T = 17.26 ^oC

6 0
3 years ago
An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu
Maru [420]

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

       = e × V

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Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

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4 0
2 years ago
Which of the following are density labels? <br> a. Kg/L <br> b. g/m <br> c. g/mL <br> d. cm/g
bearhunter [10]
Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance.<span> </span><span>It has standard units of kg/m^3 or g/mL. So, the best answer is option C.</span>

4 0
3 years ago
Read 2 more answers
A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the vol
andrezito [222]

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling C_0 the capacitance of the capacitor in air, the charge Q, the capacitance C_0 and the voltage (V_0=391 V) are related by

C_0 =\frac{Q}{V_0} (1)


when the source is disconnected the charge Q remains on the capacitor.


When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

C=k C_0 = 5.4 C_0

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

V=\frac{Q}{C}=\frac{Q}{5.4 C_0}

And since Q=C_0 V_0, substituting into the previous equation, we find:

V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V



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Hopefully this will help you.

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