A device that has the capacity to receive and store electrical energy is a(n) <br> .

In most cultures, the difference between the majority viewpoint and the minority viewpoint is that the majority group is made to

Maru [420]

If I've understood it correctly, the right one should look like this: In most cultures, the difference between the majority viewpoint and the minority viewpoint is that the majority group is made to feel normal and the minority group is made to feel divergent.

6 0

3 years ago

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A 3500 kg truck is parked on a 7.0∘ slope. How big is the friction force on the truck?

Arada [10]

Answer:

4180 N

Explanation:

In order to keep the truck balanced and stationary, the force of friction on the truck must be equal to the component of the weight of the truck acting parallel to the slope.

The component of the weight of the truck acting parallel to the slope is:

$mg sin \theta$

where

m = 3500 kg is the mass of the truck

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=7.0^{\circ}$ is the angle of the slope

Therefore, the force of friction must be equal to

$F=(3500)(9.8)(sin 7.0^{\circ})=4180 N$

6 0

3 years ago

Read 2 more answers

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.15 s later. Ignor

pogonyaev

Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

$y =- \frac{1}{2} gt^2 + v_0 t + y_0$

where:

acceleration g = 9.8 m/s²

time t

initial velocity v₀

initial height y₀

a) lets divide (a) in two parts:

1.part: How long will it take the second ball to fall down?

$v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}$

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

$y = 0, y_0 = 19.6, t = \sqrt{\frac{2y_0}{g}} + 1.15\\ 0 =- \frac{1}{2} gt^2 + v_0t + y_0$

This leaves only one unknown: v₀

$v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}$

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

$t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2$

and how fast will it go:

$v' = -gt + v_0$

2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

$0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2$

Now let's use this result in the equation for the first ball:

$0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m$

3 0

3 years ago

A new ride being built at an amusement park includes a vertical drop of 126.5 metered. Starting from rest, the ride vertically d

Delvig [45]

Answer:

The amount of Potential Energy lost in the form of Thermal Energy is equal to 41.64 MJ.

Explanation:

mass of the cart and passengers = m = 3.5*10⁴ kg

Height = h = 126.5 m

g = 9.8 ms⁻²

v = 10 ms⁻¹

At the top position the Total energy of the ride is in the form of potential energy given as:

P.E = mgh

= 3.5*10⁴*9.8*126.5

= 43389500 J

At the bottom of the drop, all the Potential Energy will be converted into K.E as the ride develops speed given as:

K.E = 0.5*m*v²

= 0.5*3.5 x 10⁴*100

= 1750000 J

The K.E is much less than the P.E energy even though all the P.E of the ride has been converted. The loss of energy is due to the formation of thermal energy which can be calculated as:

Thermal Energy lost = Total energy in the form of P.E(top) - Total energy in the form of K.E(bottom)

= 43389500 - 1750000

= 41639500 J

= 41.64 M J approx M J = mega joule = 10⁶ J

7 0

3 years ago

Choose all the answers apply Technology______.

Ahat [919]

I’m sure the answer should be E, because the rest don’t make a lot of sense the way that they’re stated

7 0

3 years ago

A device that has the capacity to receive and store electrical energy is a(n) .