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3 years ago

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A 500.-kg roller coaster car starts from rest at the top of a 60.0-meter hill.

1 answer:

Paraphin [41]3 years ago

8 0

1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.

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a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago

A simple motor converts ________energy into________energy.

k0ka [10]

I think it is C. I hope I helped.

8 0

3 years ago

Read 2 more answers

What is the current in a 120V circuit if the resistance is 20Ω?

Kaylis [27]

We have: $I=\frac{U}{R}=\frac{120}{20}=6A$

ok done. Thank to me :>

6 0

2 years ago

A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat

forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, $a=9.8\ m/s^2$

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

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

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

$F=30\times (9.8-\dfrac{(7.3)^2}{15})$

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

5 0

3 years ago

A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat

KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

$L_s$ = Latent heat of steam

$s_w$ = Specific heat of water

$L_i$ = Latent heat of ice

v = Velocity of ice

$\Delta T$ = Change in temperature

Amount of heat required for steam

$Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)$

Heat released from water at 100 °C

$Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6$

Heat released from water at 0 °C

$Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)$

Total heat released is

$Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m$

The kinetic energy of the bullet will balance the heat

$K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s$

The velocity of the ice would be 2452.79432 m/s

6 0

4 years ago