Work and power!

I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?

Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

$\frac{1}{2} m {v}^{2} = mgh$

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

$\sin(15) = \frac{h}{d} \\ or \: h = d \sin(15)$

Plugging this into the energy conservation equation and cancelling m, we get

${v}^{2} = 2gd \sin(15)$

Solving for d,

$d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m$

3 0

3 years ago

Give reason for the following:

Zigmanuir [339]

a) An inflated balloon was pressed against a wall after it has been rubbed with a piece of synthetic cloth. It was found that the balloon sticks to the wall. This is because a positive and negative electric charge is produced, therefore the balloon sticks to the wall.

b) When an object is thrown up, it comes back to ground because of gravitational attraction force of earth.

c) Mountaineers suffer nose bleeding at higher altitudes because the oxygen level decreases with increase in altitude, which the body cannot adjust.

d) Foundations of high rise buildings are kept wide because more is the area of contact, less is the pressure efforts. So, foundations are wide so as to decrease the possibility of the building from falling down.

e) Deep sea divers or high altitude fliers wear special suits so as prevent their body from being crushed by the water pressure. Since water pressure is maximum at deep seas and oceans, therefore, more is the risk of being injured.

f) Walls of a dam are thickened near the base so that the dam can handle the kinetic energy pressure and prevent itself from breaking down, which if not, can lead to flooding.

HOPE IT HELPS...

5 0

2 years ago

Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.

Gelneren [198K]

The average speed between 0 h and 2.340 h is 6.97 Km/h

Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\$

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:

Total time = 2.340 – 0 = 2.340 h
Total distance = 16.3 – 0 = 16.3 Km
Average speed =?

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\Average \: speed = \frac{16.3}{2.340} \\ \\ Average \: speed = 6.97 \: Km/h \\ \\$

Learn more about average speed: brainly.com/question/24884027

8 0

2 years ago

Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to bod

boyakko [2]

Answer:

The total amount of heat needed will be $Q_T=21.411kcal$ .

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

$m=183g$

$l_f=80\frac{cal}{g} =334\frac{J}{g}$

$l_w=1\frac{cal}{g} =4.184\frac{J}{g}$

i) The fusion heat will be:

$Q_f=l_fm=14640cal=14.640kcal$

ii) The heat needed to warm the water from $T_i=0^{\circ}C$ to $T_i=37^{\circ}C$ will be:

$Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal$

So, the total amount needed will be the sum of these two results:

$Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal$ .

8 0

3 years ago

Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.

Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let $v_1$ and $v_2$ be the speed of the ball1 and the ball2 respectively, such that

$v_1>v_2\;\cdots(i)$

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, $v$ , is $\frac 12 m v^2$

and the potential energy is due to the change in height is $mgh$ [where $g$ is the acceleration due to gravity]

So, the total energy of ball1,

$=\frac 12 m v_1^2 + mgh\;\cdots(ii)$

and the total energy of ball1,

$=\frac 12 m v_2^2 + mgh\;\cdots(iii)$ .

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, $\frac 12 m v_1^2 >\frac 12 m v_2^2$

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0

2 years ago