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3 years ago

NEED ASAP PLEASE !!

Physics

2 answers:

Advocard [28]3 years ago

8 0

Answer:

Equal to the angle of reflection.

Explanation:

There are two laws given for the reflection through mirrors. According to the law of reflection :

1. The angle of incidence, the angle of reflection, the incident ray, the reflected ray and the point of incidence all lie in the same plane.

2. The angle of incidence is equal to the angle of reflection.

So, when light is reflected by a mirror, the angle of incidence is always equal to the angle of reflection. Hence, the correct option is (c).

Lady bird [3.3K]3 years ago

3 0

I believe the correct response would be C. Equal to the angle of reflection.

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A 1kg cart slams into a stationary 1kg cart at 2 m/s. The carts stick together and move forward at a speed of 1 m/sl. Determine

finlep [7]

Answer:

No, it is not conserved

Explanation:

Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.

The total kinetic energy before the collision is:

$K_i = K_1 + K_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2=\frac{1}{2}(1 kg)(2 m/s)^2+\frac{1}{2}(1 kg)(0)^2=2 J$

where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.

After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is

$K_f = \frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(1 kg+1kg)(1 m/s)^2=1 J$

So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.

3 0

4 years ago

A cyclist travels 1850 m north and then 1200 m south. Find both the distance it has traveled and the magnitude of its displaceme

finlep [7]

Answer:

The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.

Explanation:

Distance refers to the length between any two points in space, while displacement refers to the distance from a start position to an end position regardless of the path.

In other words, distance refers to how much space an object travels during its movement; is the quantity moved. It is also said to be the sum of the distances traveled. The distance traveled by a mobile is the length of its trajectory and it is a scalar quantity. In this case, the distance is calculated as:

1850 m + 1200 m= 3,050 m

Displacement refers to the distance and direction of the final position from the initial position of an object. The displacement effected is a vector quantity. The vector representing the displacement has its origin in the initial position, its end in the final position, and its module is the distance in a straight line between the initial and final positions. That is, when expressing the displacement it is done in terms of the magnitude with its respective unit of measurement and the direction because the displacement is a vector type quantity. Mathematically, the displacement (Δd) is calculated as:

Δd= df - di

where df is the final position and di is the initial position of the object.

In this case, the displacement is calculated as:

1850 m - 1200 m= 650 m

Since the distance to the north is greater, the direction of travel will be to the north.

<u><em>The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.</em></u>

6 0

3 years ago

You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5 + 0.6 × 20c = 343.5

so wavelength of sound λ = c / f

wavelength = 343.5 / 686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............

for d = 0.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)$ = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)$ = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)$ = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475 = 2.975 m

distance 2 = 2.5 + 3.125 = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0

4 years ago

A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on

Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

$F_f = ma$ (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

$m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg$

And we can fidn the acceleration by using the formula:

$a=\frac{v-u}{t}$

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

$a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2$

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

$F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N$

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0

3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago