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2 years ago

15

To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo

u choose to throw a rubber ball or a beanbag of equal size and weight?

1 answer:

Setler [38]2 years ago

8 0

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

$P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\ P = -2mv_{i}$

For Beanbag

value of impulse will large if velocity is zero.

$v_{f}=0\\ P = -mv_{i}$

Explanation:

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A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv

zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 $KV_{LL}$

(c) The sending-end voltage for 0.9 leading power factor is 33.40 $KV_{LL}$

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0

2 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

8 0

2 years ago

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

6 0

2 years ago

The van of Hans and Frans is stuck on slippery ice, so they must get out and move it by hand. Hans pushes

Irina18 [472]

Answer: a= ff+fh/m

Explanation: bc khan academy said it was d. a=ff +fh/m

4 0

2 years ago

Suppose you have 600.0 grams of room temperature water (20.0 degrees Celsius) in a thermos. You drop 90.0 grams of ice at 0.00 d

IrinaK [193]

Answer:

$T_{f}$ = 7.02 ° C

Explanation:

The liquid water gives heat to melt the ice (Q₁) maintaining the temperature of 0 ° C and then the two waters are equilibrated to a final temperature.

Let's start by calculating the heat needed to melt the ice

Q₁ = m L

Q₁ = 0.090 3.33 10⁵

Q₁ = 2997 10⁴ J

This is the heat needed to melt all the ice

Now let's calculate at what temperature the water reaches when it releases this heat

Q = M $c_{e}$ (T₀ - $T_{f}$ )

Q₁ = Q

$T_{f}$ = T₀ - Q₁ / M $c_{e}$

$T_{f}$ = 20.0 - 2997 104 / (0.600 4186)

$T_{f}$ = 20.0 - 11.93

$T_{f}$ = 8.07 ° C

This is the temperature of the water when all the ice is melted

Now the two bodies of water exchange heat until they reach an equilibrium temperature

Temperatures are

Water of greater mass T₀₂ = 8.07ºC

Melted ice T₀₁ = 0ºC

M $c_{e}$ (T₀₂ - $T_{f}$ ) = m $c_{e}$ ( $T_{f}$ - T₀₁)

M T₀₂ + m T₀₁ = m $T_{f}$ + M $T_{f}$

$T_{f}$ = (M T₀₂ + 0) / (m + M)

$T_{f}$ = M / (m + M) T₀₂

let's calculate

$T_{f}$ = 0.600 / (0.600 + 0.090) 8.07

$T_{f}$ = 7.02 ° C

4 0

2 years ago