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3 years ago

Energy needs, in total kilocalories per day, are greater during _____________ than any other time of life.

Physics

2 answers:

frutty [35]3 years ago

6 0

Answer:

b. adolescence

Explanation:

malfutka [58]3 years ago

5 0

Energy needs, in total kilocalories per day, are greater during adolescence than any other time of life. The correct option among all the options that are given in the question is the second option or option "b". Only during the time of pregnancy will a girl require more kilocalories per day than during the time of adolescence.

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A wave carries _____ from one place to another Mechanical waves carry energy through ______

gulaghasi [49]

A wave carries energy from one place to another.

mechanical waves carry energy through MEDIUM.

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3 years ago

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What is the speed of a car that travels 50m in 25min? *

ArbitrLikvidat [17]

Answer:

the speed is 2

Explanation:

speed= distance/ time

speed= 50/25

= 2

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2 years ago

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If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?

Alisiya [41]

Answer:

The spring constant is 3750 N/m

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m

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3 years ago

An astronaut goes to Mars to do some experiments. Explain why her mass stays the same but her weight changes.

Snowcat [4.5K]

Because mass does not change from place to place but weight does change from place to place... why? because weight is the amount of gravitational force on an object and mass is the amount of matter in an object. mars has less gravitational force so an object will weigh less than it really weighs there

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2 years ago

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago