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3 years ago

Calculate the magnitude of the total impulse applied to the car to bring it to rest

Physics

1 answer:

n200080 [17]3 years ago

5 0

Answer:

Total impulse = $mv$ = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

$|J|=|p_f-p_i|$

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

$p_i=mv$

The final momentum of the car is given as:

$p_f=m(0)=0$

Therefore, the impulse is given as:

$|J|=|p_f-p_i|=|0-mv|=|-mv|=mv$

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

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Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.

Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

$ma = \text{Mass}\times \text{Acceleration}$

$ma = kg \cdot \frac{m}{s^2}$

At the same time we have that

$\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2$

$\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2$

$\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}$

Therefore there is not have same units and both are not consistent and the correct answer is B.

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3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

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3 years ago

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weeeeeb [17]

Answer: Solubility

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3 years ago

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr

Agata [3.3K]

Answer:

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