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3 years ago

What does the term “energy crisis” mean?

1 answer:

5 0

An energy crisis<span> is any significant (bottleneck; logistics; or price rise) in the supply of energy resources to an economy. In popular literature, it often refers to one of the energy sources used at a certain time and place, in particular those that supply national electricity grids or those used as fuel in vehicles.</span>

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A 9.0 cm object is 3.0 cm from a lens, which has a focal length of –12.0 cm.

artcher [175]

The answers are -2.4 for the distance. 7.2 for the height. Concave for the type

7 0

2 years ago

Read 2 more answers

A train travels 83 kilometers in 2 hours, and then 79 kilometers in 2 hours. What is its average speed?

snow_tiger [21]

If you could please give me a already given speed I could estimate it. since there is no speed shown you wouldn't be able to estimate the speed of the moving train.

7 0

2 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

2 years ago

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

2 years ago

As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile

Ber [7]

Answer:

a) $\Delta T= 100^{\circ}C$

b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

c) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

d) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

$T=(d+30) ^{\circ}C$ ...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

$\Delta T= T_f-T_i$ ..............................(2)

where:

$T_f$ = final temperature

$T_i$ = initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

$\bigtriangledown T=\frac{\Delta T}{\Delta d}$ ......................(3)

where:

$\Delta d$ = change in distance

$\bigtriangledown T=$ change in temperature with respect to distance

putting the respective values in eq. (3)

$\bigtriangledown T=\frac{100}{100}$

$\bigtriangledown T=1^{\circ}C.mile^{-1}$

(c)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

(d)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

4 0

2 years ago