In 10 seconds a car accelerates 4m/s^2 to 50

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is 55 mi/h

so if he drives all the to the meeting with max speed then it takes $=\frac{450}{55}=8.181 h$

and total allowable time is 10.8

Therefore longest time he can spend over dinner is $10.8-8.181 \approx 2.6 hours$

5 0

3 years ago

Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 m from the b

EleoNora [17]

Answer:

0.317 s

Explanation:

distance of fan A = 18.3 m

distance of fan B = 127 m

speed of sound (s) = 343 m/s

What is the time difference between hearing the sound at the two locations?

time (T) = distance / speed

time for sound to reach fan A = 18.3 / 343 = 0.053 s

time it takes for sound to reach fan B = 127 / 343 = 0.370 s
time difference = 0.370 - 0.053 = 0.317 s

5 0

3 years ago

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

levacccp [35]

Answer:

$Q = 8,345 * v$

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

$Q = 8,345 * v$

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

8 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago

Which energy transformations occur when a candle burns

gayaneshka [121]

The heat from the wick melts the wax which gets absorbed in the wick and then gets burnt (which is really oxidation) to produce heat energy as well as light energy. The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs, and produces heat and light.

6 0

3 years ago

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