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3 years ago

Mr.maxwell, an english teacher, thinks all the sudents in the class who play sports are lazy

Physics

2 answers:

aivan3 [116]3 years ago

8 0

What’s the question supposed to be

ValentinkaMS [17]3 years ago

4 0

Slap em Bc they not supposed to me sleeping in her class and that’s on period

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How much power is needed to push a 95 kg chest at 0.67 m/s along a horizontal floor where the coefficient of friction is 0.77?

Savatey [412]

$F = mg\mu \\ \\ P = \frac{Fdx}{dt} = Fv \\ \\ P = mg\mu\ v$

8 0

4 years ago

Trace fossils are much more common than body fossils. Which of the following best explains why trace fossils are more common?

Anika [276]

<span>the remains or impression of a prehistoric organism preserved in petrified form or as a mold or cast in rock.</span>

7 0

3 years ago

A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration

Oksi-84 [34.3K]

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

Therefore, the acceleration of the box is 3 m/s²

7 0

3 years ago

At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 71.9 V/m. Fi

kirza4 [7]

Answer:

a) Magnetic field strength, B = 2.397 * 10⁻⁷ T

b) Total energy density, U = 4.58 * 10⁻⁸ J/m³

c) Power flow per unit area, S = 13.71 W/m²

Explanation:

a) Electric field strength, E = 71.9 V/m

The relationship between the Electric field strength and the magnetic field strength in vacuum is:

E = Bc where c = 3.0 * 10⁸ m/s

71.9 = B * 3.0 * 10⁸

B = 71.9 / (3.0 * 10⁸)

B = 23.97 * 10⁻⁸

B = 2.397 * 10⁻⁷ T

b) Total Energy Density:

$U = \frac{1}{2} \epsilon_0E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \\U = \frac{1}{2}* 8.85 * 10^{-12}*71.9^2 + \frac{1}{2} \frac{(2.397*10^{-7})^2}{4\pi*10^{-7}}\\U = 2.29 * 10^{-8} + 2.29 * 10^{-8}\\U = 4.58 * 10^{-8} J/m^3$

c)Power flow per unit area

$S = \frac{1}{\mu_0} EB\\S = \frac{1}{4\pi * 10^{-7} } * 71.9 * 2.397 * 10^{-7}\\S = 13.71 W/m^2$

6 0

3 years ago

A ball with a weight of 2 N is attached to the end of a cord of length 2m . The ball is whirl in a vertical circle counterclockw

ZanzabumX [31]

Answer:

we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

Explanation:

Weight of the ball is given as

$W = 2N$

so we have

$m = \frac{W}{g}$

$m = 0.204 kg$

now tension force at the top is given as

$T_{top} = 7 N$

$T_{bottom} = 15 N$

Now at the top position by force equation we can say that ball will have two downwards forces

1) Tension force

2) Weight of the ball

so net force on the ball is given as

$F_{net} = T + W$

$F_{net} = 7 + 2 = 9 N$

So we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

8 0

3 years ago