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2 years ago

Explain an amusement park you've ridden discuss the conversion of energy on that ride

1 answer:

6 0

Roller coaster...

going up... building the potential energy....

SCREAM... AND SCREAM... kinetic energy upon kinetic energy being used

rinse and repeat

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A vw bug is driving north on a straight two lane road at a constant 54 mph and chevy impala south at 47 on the same road (but di

mario62 [17]

Velocity of vw bug is given as

$v_1 = 54 mpH$ towards north

velocity of imphala is given as

$v_2 = 47 mph$ towards south

now we need to find velocity of imphala relative to bug

$v_{21} = v_2 - v_1$

$v_{21} = 47 - (-54) = 101 mpH$

so the speed is 101 mpH towards south so correct option is C

5 0

2 years ago

Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 25°, and Paul is pulling a large crate u

Alisiya [41]

Explanation:

It is given that,

The ramp is tilted upwards at 25 degrees and Paul is pulling a large crate up the ramp with a rope that angles 10° above the ramp.

Total angle with respect to ramp is 35 degrees.

If Paul pulls with a force of 550 N.

The horizontal component of the force is given by :

$F_x=F\ cos\theta$

$F_x=550\ cos(35)$

$F_x=450.53\ N$

The vertical component of the force is given by :

$F_y=F\ sin\theta$

$F_y=550\ sin(35)$

$F_y=315.46\ N$

Hence, this is the required solution.

4 0

2 years ago

By how much does the pressure of a gas in a rigid vessel decrease when the temperature drops from 0 ∘C to –1∘C?

Ugo [173]

When a problem says a rigid vessel, it means that volume is constant. At constant V, pressure and temperature are indirectly proportional. We calculate as follows:

P1/T1 = P2/T2
P1/P2 = T1/T2
P1/P2 = 273.15 / 272.15
P1/P2 = 1.00

Hope this helps. Have a nice day.

7 0

3 years ago

A bowling ball has a mass of 5 kg. A student rolls the bowling ball one time at 1 m/s and a second time and 2 m/s. Compare the m

sasho [114]

The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls

5 0

2 years ago

An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp

Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

$k=\dfrac{1}{2}mv^2$

$k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2$

k = 1150 J

In two significant figure, $k=1.2\times 10^3\ J$

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. $P=\dfrac{W}{t}=\dfrac{\Delta K}{t}$

$P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}$

P = 115 watts

In two significant figures, $P=1.2\times 10^2\ Watts$

Hence, this is the required solution.

6 0

2 years ago