Why is space black if the sun is shining

Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coef

Alex777 [14]

A. The acceleration during the slide is 6.86 m/s²

B. The time taken to slide until he stops is 1.2 s

<h3>How to determine the force of friction</h3>

Mass (m) = 81.5 Kg
Coefficient of friction (μ) = 0.7
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
Frictional force (F) =?

F = μN

F = 0.7 × 798.7

F = 559.09 N

<h3>A. How to determine the acceleration</h3>

Mass (m) = 81.5 Kg
Frictional force (F) = 559.09 N
Acceleration (a) =?

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

<h3>B. How to determine the time </h3>

Initial velocity (u) = 8.23 m/s
Final velocity (v) = 0 m/s
Decceleration (a) = -6.86 m/s²
Time (t) =?

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

Learn more about acceleration:

brainly.com/question/491732

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4 0

2 years ago

an always be used to calculate the electric field. relates the electric field at points on a closed surface to the net charge en

Temka [501]

Complete Question:

Gauss's law:

Group of answer choices

A. can always be used to calculate the electric field.

B. relates the electric field throughout space to the charges distributed through that space.

C. only applies to point charges.

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

E. relates the surface charge density to the electric field.

Answer:

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

8 0

2 years ago

The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo

spin [16.1K]

Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

$W=\vec F\ \vec r$

It can be written in its scalar version as

$W=F.d.cos\theta$

Being F and d the magnitudes of the force and displacement, and $\theta$ the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

$W=N.d.cos90^o=0$

The work is zero because the force and the displacement are perpendicular

The static friction force exists only when the object is being applied a force of a magnitude not large enough to produce movement, i.e. the object is at rest. If the object is moved, the friction force is still present, but it's called dynamic friction force, usually smaller than the static.

Since in this case, there is no displacement, d=0, and the work is

$W=F_r(0)cos180^o=0$

3 0

3 years ago

A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga

BartSMP [9]

Answer:

Total work done in expansion will be $3.60\times 10^5J$

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm $=1.01\times 10^5Pa$

So 2.10 atm $=2.10\times 1.01\times 10^5=2.121\times 10^5Pa$

Volume is increases from 3370 liter to 5.40 liter

So initial volume $V_1=3.70liter$

And final volume $V_2=5.40liter$

So change in volume $dV=5.40-3.70=1.70liter$

For isobaric process work done is equal to $W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J$

So total work done in expansion will be $3.60\times 10^5J$

5 0

2 years ago

Briefly explain the two postulates of relativity, how they differed from classical physics, and why Einstein was led to propose

Ray Of Light [21]

Explanation:

The two postulates of special theory of relativity

Postulate 1: The law of physics are invariant under any of inertial frame of reference.

Postulate 2: The velocity of light is remains same in each ans every frame of reference and independent of relativity.

They are differ from classical mechanics that in classical mechanics there is no change in mass and length in relative velocity but in relativistic mechanics it changes.

These two postulates implements in phenomenon like time dilation , length contraction etc.

Thanks

3 0

3 years ago