A waiter is carrying a tray above his head and walking at a constant velocity. If he applies a force of 5.0 newtons on the tray and covers a distance of 10.0 meters, how much is the work being done?

Answers

Answer 1
Answer: w = f × x = 5 × 10 = 50
Answer 2
Answer:

the other answer is wrong, the answer is A. 0 joules for plato users, i got 100 on the test


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Which of these are terrestrial planets? Select all that apply.a. Mars
b. Earth
c. Venus
d. Jupiter E. Mercury F. Neptune

Answers

The answers to the question above would be letters a, b, c, and d. The terrestrial planets are those that are close to the sun, namely, Mercury, Venus, Earth, and Mars. The others are known as the gas giants since they have a thick atmosphere and are considerably cold.

Answer:

a. Mars

b. Earth

c. Venus

Explanation:

Terrestrial planets are defined as planets which are formed by silicate. An important characteristic is that these planets have a solid surface. On the other hand, there are gaseous planets, which are bigger and more separated of the Sun.

In our Solar system, there are 4 terrestrial planets, which are Mercury, Venus, Earth, and Mars. Notice that they are the inner planets of the system.

So, in this case, the right chice is a. Mars, b. Earth and c. Venus. All three are terrestrial planets, that is, they have a rock solid surface.

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3. What is the density of gold

Answers

The density of the sample of gold whose mass is 38.6 grams and with a volume of 2 cm³ is 19300 kg/m

Density: This can be defined as the ratio of mass and velocity. The S.I unit of density is kg/m³.

The question above can be solved using the formula below

D = m/V.................... Equation 1

Where D = density of the sample of gold, m = mass of gold, V = volume of gold.

From the question,

Given: m = 38.6 g, = 0.0386 kg, V = 2 cm³ = 0.000002 m³

Substitute these values into equation 1

D = 0.0386/0.000002

D = 19300 kg/m³

Hence the density of the sample of gold is 19300 kg/m³

Learn more about density here: brainly.com/question/23562510

Answer:

19.3 g/cm^3

Explanation:

The density of a material is given by

\rho = (m)/(V)

where

m is the mass of the sample

V is its volume

for the sample of gold in this problem,

m = 38.6 g

V=2 cm^3

Substituting into the formula, we find the density of gold:

\rho = (38.6)/(2)=19.3 g/cm^3

Water is a nonpolar molecule true or false​

Answers

Answer:

Water is non-polar molecule False

Water is polar molecule because the electronegativity of Oxygen is much greater than electronegativity of hydrogen and it has also a bend shape that is why it is polar molecule.

Explanation:

I hope this will help you :)

False. Water is a polar molecule.

About water

Water (H2O) consists of two hydrogen atoms and one oxygen atom. The oxygen atom has a higher electronegativity, meaning it attracts electrons more strongly than hydrogen.

As a result, the shared electrons in the covalent bonds between hydrogen and oxygen are pulled closer to the oxygen atom, creating an uneven distribution of charge.

This leads to a partial negative charge (δ-) on the oxygen atom and partial positive charges (δ+) on the hydrogen atoms, making water a polar molecule.

More on polarity of water can be found here: brainly.com/question/30530422

#SPJ6

Which of the following bodies has a brightness that surpasses Venus’? My options are Jupiter, Mars, the stars, and the Moon. Jupiter and mars are obviously out and Venus is brighter than all the stars in the sky (besides the sun lol) so it would be the moon right? just a lil confused.

btw i'm not sure what subject to post this in since there's no astronomy subject.

Answers

Answer: the Moon.

Explanation:

Brightness of celestial object is given in terms of magnitude. Apparent magnitude reflects how bright an object appears from the Earth. The apparent magnitude of the Moon is (-12.7 )greater than the apparent magnitude of the Venus (-4.4). The more negative value, the brighter is the object. Thus, the brightness of the moon surpasses the brightness of Venus.

Hi there!

The answer would be the stars. Since the sun is a star, the star is the answer. A star is much, much hotter than Venus, making it much brighter. 

Hope this helps!

A coin is placed 32 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When the speed of the coin is 110 cm/s (rotating at a constant rate), the coin just begins to slip. The acceleration of gravity is 980 cm/s^2 . What is the coefficient of static friction between the coin and the turntable?

Answers

Answer:

0.39

Explanation:

distance from the center (r) = 32 cm = 0.32 m

speed of the coin (v) = 110 cm/s = 1.1 m/s

acceleration due to gravity (g) = 980 m/s^{2} = 9.8 m/s^{2}

find the coefficient of static friction (k) between the coin and the turn table

frictional force = kmg

before the table begins to move, the frictional force balances the centripetal force ((mv^(2) )/(r))

therefore

frictional force = centripetal force

kmg = (mv^(2) )/(r)

kg = (v^(2) )/(r)

k = (v^(2) )/(r) ÷ g

k = (1.1^(2) )/(0.32) ÷ 9.8 = 0.39

A stone with a mass of 0.70 kg is attached to one end of a string 0.80 m long. The string will break if its tension exceeds 65.0 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. Find the maximum speed the stone can attain without breaking the string.

Answers

Answer:

The maximum speed the stone can attain without breaking the string is 8.62 m/s .

Explanation:

Given :

Mass of stone , m = 0.7 kg .

Length of string , l = 0.8 m .

It is also given that the stone will break if its tension exceeds 65.0 N.

Now , we know tension in the rope due to rotation is equal to the centripetal acceleration .

Therefore , the maximum speed the stone can attain without breaking the string is less than or equal to 65 N .

So , (mv^2)/(r)=65

Putting all value in above equation we get :

(0.7* v^2)/(0.8)=65\n\nv= 8.62\ m/s

Therefore , maximum speed the stone can attain without breaking the string is 8.62 m/s .