Ask question

Ask question

All categories

ArbitrLikvidat [17]

2 years ago

12

Albert Jr. takes his new skateboard out to the top of the loading ramp. The ramp is 30 m long and sloped at an angle of 15 degre

es relative to the horizontal. Assuming he starts from rest, and neglecting friction, about how fast is Albert Jr. going at the bottom of the ramp?

1 answer:

WINSTONCH [101]2 years ago

5 0

Answer:

vf = 12.343 m/s

Explanation:

Given:

- The Length of the ramp s = 30 m

- The angle of the ramp θ = 15 degrees

- Initial velocity at the top of ramp vi = 0

- Neglect Friction

Find:

How fast is Albert Jr. going at the bottom of the ramp?

Solution:

- We can use conservation of energy for the skateboard at top and bottom of the ramp.

ΔK.E = ΔP.E

Where, ΔK.E : Change in kinetic energy

ΔP.E : Change in gravitational potential energy

0.5*m*(vf^2 - vi^2) = m*g*Δh

Where, m: The mass of the object

Δh: Change in elevation from top to bottom

Δh = s*sin(θ)

- Substitute and simplify:

0.5*(vf^2 - vi^2) = g*s*sin(θ)

vf^2 = 2*g*s*sin(θ)

vf = √(2*g*s*sin(θ)) = √(2*9.81*30*sin(15))

vf = 12.343 m/s

- The velocity at the bottom of ramp would be 12.343 m/s

You might be interested in

A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee

Olegator [25]

Quite low

Gravitation force = $\frac{6.67* 10^{-11}*0.145*6.8 }{ 0.5^{2} }$ = 26.3 * $10^{-11}$ = 2.63 * $10^{-10}$ N

6 0

3 years ago

Read 2 more answers

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago

What is the period of a sound wave having a frequency of 340 Hz

sertanlavr [38]

Answer:

lambda = 343 m/s divided by 340 Hz = 1.009 seconds

Hope it helps and have a wonderful day!

4 0

3 years ago

Read 2 more answers

In your own words, explain what a field is in science. Use specific examples to support your explanation.

tresset_1 [31]

Answer:

Field, In physics, a region in which each point is affected by a force.

Explanation:

4 0

3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0

3 years ago

Read 2 more answers