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3 years ago

A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the

1 answer:

8 0

The energy of the ski lift at the base is kinetic energy:
$K= \frac{1}{2}mv^2$
where m is the mass of the ski lift+the people carried, and $v=15.5 m/s$ is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the height at the top of the hill.

So, since the total energy must conserve, we have
$U=K$
and so
$mgh = \frac{1}{2}mv^2$
from which we find the height:
$h= \frac{v^2}{2g}= \frac{(15.5m/s)^2}{2\cdot 9.81 m/s^2}=12 m$

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Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

Second stone

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago

Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How

grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object ( $\Delta s$ ), in meters, can be determined by the following expression:

$\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2}$ (1)

Where:

$v_{o}$ - Initial velocity, in meters per second.

$t$ - Time, in seconds.

$a$ - Acceleration, in meters per square second.

If we know that $v_{o} = 5\,\frac{m}{s}$ , $t = 6\,s$ and $a = 2\,\frac{m}{s^{2}}$ , then the box displacement after 6 seconds is:

$\Delta s = 66\,m$

The box displacement after 6 seconds is 66 meters.

5 0

3 years ago

How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules

dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0

2 years ago

What 3 points must an object possess to be considered a planet

Vitek1552 [10]

1. It must be an object which independently orbits the Sun (this means moons can't be considered planets, since they orbit planets)
2. It must have enough mass that its own gravity pulls it into a spheroidal shape.
3. It must be large enough to "dominate" its orbit (i.e. its mass must be much larger than anything else which crosses its orbit).

4 0

2 years ago

A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

JulsSmile [24]

$m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}$

7 0

2 years ago