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3 years ago

15

You have been asked to make a roller coaster more exciting. The owners want the speed at the bottom of the first hill doubled. H

ow much higher must the first hill be built?

1 answer:

qaws [65]3 years ago

4 0

Answer:

The height will be 4 times.

Explanation:

Given that,

The speed at the bottom of the hill doubled.

We need to calculate the height

Using conservation of energy

$K.E_{t}+P.E_{t}=K.E_{b}+P.E_{b}$

$K.E_{b}=P.E_{t}$

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

$\dfrac{1}{2}m(4v^2)=mgh$

Therefore,

$P.E = mg(4h)$

Here, m and g are constant

Hence, The height will be 4 times.

You might be interested in

What is the impulse

vichka [17]

Impulse is the integral of a force, F.
Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!

5 0

3 years ago

When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these

Zigmanuir [339]

Answer:

a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

v = $\sqrt{ \frac{T}{ \mu } }$

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

m = 2m₀

let's find the new linear density

μ = m / l

iinitial density

μ₀ = m₀ / l

final density

μ = 2m₀ / lo

μ = 2 μ₀

we substitute in the equation for the velocity

initial v₀ = $\sqrt{ \frac{T_o}{ \mu_o} }$

with the new dough

v = $\sqrt{ \frac{T_o}{ 2 \mu_o} }$

v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }

v = 1 /√2 v₀

v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

μ = μ₀

in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

μ = 3m₀ / l₀

μ = 3 m₀/l₀

μ = 3 μ₀

v = $\sqrt{ \frac{T_o}{ 3 \mu_o} }$

v = 1 /√3 v₀

v = 0.577 v₀

d) l = 2l₀

μ = m₀ / 2l₀

μ = μ₀/ 2

v = $\sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }$

v = √2 v₀

v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

μ = 10 m₀/2l₀

μ = 5 μ₀

we look for speed

v = $\sqrt{ \frac{T_o}{5 \mu_o} }$

v = 1 /√5 v₀

v = 0.447 v₀

5 0

3 years ago

What is the radius of a circle thats the diameter of 480?

Luda [366]

The answer is 12.36. hoped this helped!

4 0

3 years ago

A ball thrown horizontally at 12.6 m/s from the roof of a building lands 20.0 m from the base of the building

S_A_V [24]

Answer:

1.59 seconds

12.3 meters

but if you are wise you will read the entire answer.

Explanation:

This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.

Time

<u>Given</u>

a = 0 This is the critical point. There is no horizontal acceleration.

d = 20 m

v = 12.6 m/s

<u>Formula</u>

d = vi * t + 1/2at^2

<u>Solution</u>

Since the acceleration is 0, the formula reduces to

d = vi * t

20 = 12.6 * t

t = 20 / 12.6

t = 1.59 seconds.

It takes 1.59 seconds to hit the ground

Height of the building

<u>Givens</u>

t = 1.59 sec

vi = 0 Another critical point. The beginning speed vertically is 0

a = 9.8 m/s^2 The acceleration is vertical.

<u>Formula</u>

d = vi*t + 1/2 a t^2

<u>Solution</u>

d = 1/2 a*t^2

d = 1/2 * 9.8 * 1.59^2

d = 12.3 meters.

The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)

The other vi is a vertical speed. It is 0.

7 0

3 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

3 years ago