In Paragraph 3 of this passage, what clues help you know the meaning of the word

Match the situation with the energy transformation ITEMBANK: Move to Top A boy shooting a rubber band across the classroom A chi

Sonbull [250]

A boy shooting a rubber band across the classroom -->
Elastic potential energy transformed into kinetic energy
The initial energy is the energy stored in the muscles of the boy's arm, which is elastic potential energy. This is converted into motion of the rubber, therefore kinetic energy

A child going down a slide on a playground --> Gravitational potential energy transformed into kinetic energy
On top of the slide, all the energy of the child is gravitational potential energy due to its height with respect to the ground (E=mgh). when it moves down the slide, this is converted into kinetic energy, because the child acquires a speed v (E=1/2 mv^2)

Rubbing your hands together to warm them on a cold day --> Kinetic energy being transformed into thermal energy 
When rubbing hands, we are moving them (kinetic energy), and this energy raises the temperature of the hand's surface (thermal energy)

Turning on a battery operated light --> 
Chemical potential energy transformed into radiant energy 
A battery works by mean of chemical reactions (chemical potential energy), producing light (so, emitting energy by radiation, i.e. radiant energy)

Using a dc electric motor --> Electrical energy transformed into kinetic energy
A dc electric motor works using currents (so, electrical energy), and the energy produced can be used for example to accelerate a car (kinetic energy)

Using a gas power heater to warm a room --> Chemical potential energy transformed into thermal energy
A gas power heater burns gases (so, chemical reaction, i.e. chemical potential energy) to raise the temperature of the room (thermal energy)

Using a hand crank generator to produce electric current --> Kinetic energy transformed into electrical energy
In a hand-crank generator, the handle is being rotated (kinetic energy) in order to produce an electric current (electrical energy)

Using the light in your room that is plugged into the wall --> Electrical energy transformed into radiant energy
The lamp works by using electrical current flowing into a resistor (electrical energy) and it produces light, so it emits energy by electromagnetic radiation (radiant energy)

3 0

3 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

4 years ago

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

timofeeve [1]

The solution would be like this for this specific problem:

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago

A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have

mixer [17]

Answer:

B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2

C) V = 0.84m/s T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

$I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2$ where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

$I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2$

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2 Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0

3 years ago

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

6 0

3 years ago