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Phoenix [80]
2 years ago
14

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

n the same direction at vTi 5 20.0 m/s (Fig. P9.22). The velocity of the car immediately after the collision is vCf 5 18.0 m/s to the east. (a) What is the velocity of the truck immediately after the colli
Physics
1 answer:
sukhopar [10]2 years ago
8 0

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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likoan [24]

Answer:

c

Explanation:

8 0
2 years ago
A school bus covers a distance of 7200 m in 1800 s. Calculate its speed.
telo118 [61]

Answer:

Speed = 4 m/sec

Explanation:

Speed = Distance / Time

Where Distance = 7200 m and Time = 1800 s

Speed = 7200 m / 1800 sec

Speed = 4 m/sec

8 0
2 years ago
A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13
tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

F = qvB

here we have

B = \frac{F}{qv}

here we know that

F = 4.8 \times 10^{-13} N

q = 1.6 \times 10^{-19} C

v = 4 \times 10^6 m/s

now from above equation we have

B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}

B = 0.75 T

8 0
2 years ago
The air within a piston equipped with a cylinder absorbs 565 J of heat and expands from an initial volume of 0.10 L to a final v
jeyben [28]

Answer:

489 J

Explanation:

According to the first law of thermodynamics:-

\Delta U = q + w

Where,

U is the internal energy

q is the heat

w is the work done

From the question,

q = + 565 J  (+ sign as the heat is being absorbed)

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,

\Delta V = 0.85 - 0.10 L = 0.75 L

P = 1.0 atm

w=-1.0\times0.75\ atmL

Also, 1 atmL = 101.3 J

So,

w=-1.0\times0.75\times 101.3\ J=-76\ J (work is done by the system)

So,

\Delta U = +565\ J-76\ J = 489\ J

5 0
2 years ago
a)A concentration C(mol/L) varies with time (min) according to the equation C=3.00exp(−2.00t) a) What are the implicit units of
telo118 [61]

Answer:

a. 3.00 must be in mol/L and 2.00 in 1/min.

b. <u>Using lineal equation:</u>

t = 0.6 min then C = 1.45 mol/L

C = 0.10 mol/L then t = 1.12 min

<u>Using exponential equation:</u>

t = 0.6 min then C = 0.90 mol/L

C = 0.10 mol/L then t = 1.70 min

Explanation:

a) C is in mol/L so, as we know that the exponential does not have units, 3.00 must be in mol/L and 2.00 in 1/min.

b) If we want to make a line between 0 an 1 we need to find the linear equation for this.

We can use the slope intercept-equation:

y = ax + b (1)

<u>Let's find a and b.</u>

a is the slope of here, so we can use the next equation to find it:

a=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

We have two points, (0,C₁) and (1,C₂)

C_{1}(0)=3.00e^{-2.00*(0)}=3.00 mol/L

C_{2}(1)=3.00e^{-2.00*(1)}=0.41 mol/L

So a will be:

a=\frac{0.41-3.00}{1-0}=-2.59

We can use the equation (1) to find b, for instance, let's choose: y_{1}=ax_{1}+b

b=y_{1}-ax_{1}=C_{1}-(2.59*0)=3.00

Finally, our linear equation will be: y=-2.59x+3.00 or  C=-2.59t+3.00

<u>Using lineal equation:</u>

t = 0.6 min then C = 1.45 mol/L

C = 0.10 mol/L then t = 1.12 min

<u>Using exponential equation:</u>

t = 0.6 min then C = 0.90 mol/L

C = 0.10 mol/L then t = 1.70 min

I hope it helps you!

5 0
2 years ago
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