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3 years ago

Which does not provide the energy to spin a turbine?

Physics

2 answers:

Sophie [7]3 years ago

6 0

It depends on what kind of the turbine to use it could be electromagnet or water in reservoir

Makovka662 [10]3 years ago

3 0

It's B, Electromagnets.

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1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?

shepuryov [24]

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N

7 0

3 years ago

A student pushes on a crate with a force of 100 Newtons directed to the right.

romanna [79]

Answer:

100N

Explanation:

Newton's third law of motion

For every action, there is an equal and opposite reaction.

Therefore 100N of force is exerted by the crate on student as a reaction to his action

6 0

3 years ago

When looking at a food label, to be considered a good source of a nutrient, the nutrient must be? A.More than 20% of your daily

aleksklad [387]

Answer:

A. More than 20% of your daily recommended amount.

Explanation:

Reading food labels can be tricky. The percent daily value listed on the right of all food labels lets you know what percent out of the recommended daily intake of each nutrient you are consuming in that specific food.

To check if the food you're consuming is a good source of that nutrient you need in higher amount, the nutrient must be labeled 20% or higher.

The rule used here is called the 5/20 rule. According to this rule, A nutrient that is 5% or below is considered less and a nutrient which is labeled 20% or higher is considered good enough in that food source.

8 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

3 years ago

10.

myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$