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2 years ago

5

Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc

ity of car B after the collision?

1 answer:

Nuetrik [128]2 years ago

4 0

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

$v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2$

The data given are;

m1 = 281kg
u1 = 2.82m/s
m2 = 209kg
u2 = -1.72m/s
v1 = ?

Let's substitute the values into the equation.

$v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s$

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.

Learn more about elastic collision here;

brainly.com/question/7694106

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Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30

barxatty [35]

For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles

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3 years ago

The process that enhances some properties of an object at the expense of other properties is called:

lidiya [134]

Answer:

The answer is: letter a, pop-out effect.

Explanation:

The "pop-out effect" is a phenomenon which allows the person's precognitive processes to detect a<em> visual stimulus that is potentially the most meaningful one</em> in a person's spatial field of attention. The pop-up effect occurs when a person distinguishes one object from the rest.

For example, when a child chooses among pictures in different colors, it is common for the child to point at colored pictures rather than grayscale pictures. This is an example of a pop-out effect. <u>The properties of the colored pictures is more preferred by the child thus, causing him not to choose or mind the grayscale images.</u>

Thus, this explains the answer.

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3 years ago

A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when

Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

<em>Throat area of the nozzle, </em> $A^*$ <em> = 10 cm² = 0.001 m²</em>
<em>The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²</em>
<em>Air pressure at sea level = 101.325 kPa</em>

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

$= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896$

From supersonic isentropic table, at $\frac{A}{A^*} = 2.896$ , we can determine the following;

$M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954$

The tank pressure is calculated as follows;

$\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa$

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn more about converging-diverging nozzle design here: brainly.com/question/13889483

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2 years ago

A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang

Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) $E=\frac{1}{2}mv^2+mgh$

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) $E=\frac{1}{2}mv^2$

The energy when the mass comes to a stop will be:

3) $E=mgh$

Setting equations 2 and 3 equal and solving for height h will give:

4) $h=\frac{v^2}{2g}$

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) $cos\phi=\frac{l-h}{l}$

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) $\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})$

8 0

4 years ago

A 28 kg mass suspends from a light rope 18 m long & is held to one side by the horizontal force, F, as shown below.

frutty [35]

Answer: 215.15 N

Explanation:

If we draw a free body diagram of the mass we will have the following:

$\sum{F_{x}}=-Tcos\theta + F=0$ (1)

$\sum{F_{y}}=Tsin\theta - mg=0$ (2)

Where $T$ is the tension force of the rope, $m=28 kg$ the mass, $g=9.8 m/s^{2}$ the acceleration due gravity and $mg$ is the weight.

On the other hand, we can calculate $\theta$ as follows:

$cos\theta=\frac{s}{l}$

$\theta=cos^{-1}(\frac{s}{l})$

Where $s=11.1 m$ and $l=18 m$

$\theta=cos^{-1}(\frac{11.1 m}{18 m})$

$\theta=51.9\°$ (3)

Now, we firstly need to find $T$ from (2):

$T=\frac{mg}{sin\theta}$ (4)

$T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}$

$T=348.69 N$ (5)

Substituting (5) in (1):

$F=Tcos\theta$ (6)

$F=348.69 N cos(51.9\°)$

Finally:

$F=215.15 N$

8 0

3 years ago