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3 years ago

Which of the following is not an example of doing work?

Physics

2 answers:

Tju [1.3M]3 years ago

5 0

Maybe holding a tray in the cafeteria line

Whitepunk [10]3 years ago

5 0

Answer:

A. Holding a tray in the cafeteria line!

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The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship

DanielleElmas [232]

Answer:

$V_m = 12.78 km/hr$

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

$F_r = \dfrac{V}{\sqrt{L g}}$

for dynamic similarities

$(\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}$

$(\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})$

$(\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})$

$V_m = 12.78 km/hr$

hence, the velocity of model will be 12.78 km/h

6 0

3 years ago

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

2 years ago

Question 1 of 25

finlep [7]

Answer:

2.753*10^-11N

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

Hence the gravitational force between the asteroid is 2.753*10^-11N

6 0

2 years ago

You accelerate a 0.43kg football 5m/s2. Calculate the force you applied to the football.

Elza [17]

The force applied on the football is 2.15 Newton.

Given the data in the question;

Mass of football; $m = 0.43kg$

Acceleration; $a = 5m/s^2$

Force applied; $F = \ ?$

To determine the force applied on the football, we use Newton's laws of motion:

$F = m * a$

Where m is the mass of the object and a is the acceleration.

We substitute our given values into the equation

$F = 0.43kg\ *\ 5m/s^2\\\\F = 2.15kg.m/s^2\\\\F = 2.15N$

Therefore, the force applied on the football is 2.15 Newton.

Learn more: brainly.com/question/2388393

7 0

2 years ago

Your friend of mass 50 kg can just barely float in fresh water.

gulaghasi [49]

The answer is 10 i think so

7 0

2 years ago