PHYSICS COLLEGE

Tell uses of cancave mirror and convex mirror.

Answers

Answer 1

Answer:

Answer:

Uses of concave mirror:

Shaving mirrors.

Head mirrors.

Ophthalmoscope.

Astronomical telescopes.

Headlights.

Solar furnaces.

Uses of convex mirror:

Convex mirrors always form images that are upright, virtual, and smaller than the actual object. They are commonly used as rear and side view mirrors in cars and as security mirrors in public buildings because they allow you to see a wider view than flat or concave mirrors.

please give me full points.

Related Questions

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3 , is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, i
WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”picture attached
An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)
As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

Answers

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

$t = (d)/(v_(max) - v_g)$

so above is the relation between all given variable

Solve for x
–30 = 5(x + 1)

Answers

Answer:

-30=5(x+1) is -7

Explanation:

distribute flip subtract 5 from both sides divide both sides by 5

-30=(5)(x)+(5)(1) (Distribute)
-30=5x+5
Flip equation
5x+5=-30
Subtract 5 from both sides
5x+5-5=-30-5
5x=-35
Divide both sides by 5
X=-7

A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH)? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?

Answers

Answer:

a) car does not skid, b) car skids, c) v = 11.07 m / s

Explanation:

a) When the car around in a curve all force must be exerted by friction, write Newton's second Law

Y axis (vertical)

N - W = 0

N = W = mg

X axis (radial

F = m a

The acceleration is centripetal

a = v² / r

fr = μ N

Let's calculate the maximum friction force

fr = μ m g

fr = 0.70 2000 9.8

fr = 13720 N

Let's calculate the force necessary to take the curve

F = m v² / r

F = 2000 11²/25

F = 9680 N

When examining these two values we see that the maximum value of the friction force is greater than the force to stay in the curve, for which the car does not skid

b) The speed of the driver is v = 18m / s, let's calculate the force to stay in the curve

F = 2000 18²/25

F = 25920 N

This force is greater than the maximum friction force, so it is a skating car

c) The friction coefficient decreases to μ = 0.5

fr = m a

μ mg = m v² / r

v = √μ g r

v = √(0.50 9.8 25)

v = 11.07 m / s

This is the maximum speed

According to the second law of thermodynamics, it is impossible for ____________. According to the second law of thermodynamics, it is impossible for ____________. heat energy to flow from a colder body to a hotter body an ideal heat engine to have the efficiency of 99% an ideal heat engine to have non-zero power. a physical process to yield more energy than what is put in

Answers

Answer:

It's impossible for an ideal heat engine to have non-zero power.

Explanation:

Option A is incomplete and so it's possible.

Option B is possible

Option D is related to the first lae and has nothing to do with the second law.

Hence, the correct option is C.

The ideal engine follows a reversible cycle albeit an infinitely slow one. If the work is being done at this infinitely slow rate, the power of such an engine is zero.

We can also stat the second law of thermodynamics in this manner;

It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat energy from a colder object to a hotter one.

This statement is known as second form or Clausius statement of the second law.

Thus, it is possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work input.

Final answer:

According to the second law of thermodynamics, it is impossible for heat energy to flow from a colder body to a hotter body, for an ideal heat engine to have an efficiency of 99%, and for a physical process to yield more energy than what is put in.

Explanation:

According to the second law of thermodynamics, it is impossible for heat energy to flow from a colder body to a hotter body. This is because heat naturally flows from a region of higher temperature to a region of lower temperature. This principle is what allows us to effectively use heat for various purposes, such as in heat engines.

An ideal heat engine is a theoretical construct used to study the efficiency of engines. The second law of thermodynamics states that no heat engine can have an efficiency of 100%, so it is impossible for an ideal heat engine to have an efficiency of 99%. This is due to the losses in heat transfer and other thermodynamic processes.

The second law of thermodynamics also implies that in any physical process, the total energy cannot increase. It is impossible for a physical process to yield more energy than what is put in. This principle is central to understanding energy conservation and the limitations of energy conversion.

Learn more about Second Law of Thermodynamics here:

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What minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration

Answers

Answer:

Coefficient of static friction = 1.84

Explanation:

Note:

Top speed = 60 mph

Acceleration of cheetah = 18 m/s²

Find:

Coefficient of static friction

Computation:

Acceleration due to gravity = 9.8 m/s²

Coefficient of static friction = Acceleration of cheetah / Acceleration due to gravity

Coefficient of static friction = 18 / 9.8

Coefficient of static friction = 1.84

9. In the graph below, what is the force being exerted onthe 16-kg cart?
A. 4N
C. 16N
B. 8N
D. 32 N

Answers

Final answer:

The force being exerted on the 16-kg cart is 32N.

Explanation:

The force being exerted on the 16-kg cart can be determined using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In the given graph, the cart is accelerating at a rate of 2 m/s2. Therefore, the force can be calculated as:

force = mass x acceleration

force = 16 kg x 2 m/s2 = 32 N

The 16-kg cart experiences a force of 32N, determined by applying Newton's second law of motion. According to this law, force equals mass multiplied by acceleration. In this scenario, the cart accelerates at a rate of 2 m/s². Substituting the values into the equation, the force exerted on the cart can be calculated as 16 kg multiplied by 2 m/s², resulting in a force of 32 N. This fundamental principle in physics establishes a quantitative relationship between force, mass, and acceleration.