Ask question

Ask question

All categories

2 years ago

7

At which of the following points does a roller coaster have the most potential energy? As it is going down a hill. At the top of

the hill. As it is climbing a hill. At the bottom of the hill.

1 answer:

vova2212 [387]2 years ago

6 0

Answer:

At the top of the hill.

Explanation:

As the roller coaster goes up the hill, kinetic energy (K.E) decreases, gravitational potential energy (G.P.E) increases .

As it reach the top of the hill, K.E becomes zero and G.P.E reaches <em>m</em><em>a</em><em>x</em><em>i</em><em>m</em><em>u</em><em>m</em> .

As it goes down the hill, K.E starts to increase and G.P.E decrease .

At the bottom of the hill, K.E reaches <em>maximum</em> and G.P.E becomes zero .

(Correct me it I am wrong)

You might be interested in

Which biome has the most extreme conditions

Lelu [443]

Well this all depends on the region you would like to know about. One biome would be The Tundra. This biome is a very bitter cold. Some times the temperature can drop to -45f! So your answer more than likely would be Tundra.

Have a wonderful day user!

3 0

2 years ago

When making sweet tea, McDonalds adds the sugar while the tea is hot. The sugar is added at this step so that the tea can be as

Lilit [14]

Answer:

d

Explanation:

we need the tea to be at first moist in order to get much more flavor. then supersaturated so that it increases the concentration of the flavor by it not being too sweet nor too salty. so just right.

3 0

2 years ago

Read 2 more answers

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

3 years ago

A glass flask whose volume is 1000 cm^3 at a temperature of 1.00Â°C is completely filled with mercury at the same temperature. W

Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is $\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = $9.18*10^{-6} \ m^3$

Increase in volume of the glass = 10⁻³ × 51.00 × $\beta _{glass}$

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = $(9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3$

$8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3$

$\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )$

$\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

Thus; the coefficient of volume expansion of the glass is $\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

5 0

2 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago