Which image illustrates reflection?
A
B
с
D

Which Image Illustrates Reflection?ABD

Answers

Answer 1

Answer: I beleive A

Explanation:

Answer 2

Answer:

A

Explanation:

We can see the light being reflected off the mirror.


Related Questions

I need help with this problem can anybody help me please , it’s physics 2 course

Answers

Answer:

ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|

Explanation:

The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:

[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]

Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is

[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]

or

[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]

What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J

Answers

Answer:

E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules

Answer:

The answer is D. 2.25 × 1017 J

Explanation:

got it right on edge 2021

The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.

Answers

Answer:

1140 J, 6840 J, 10260 J

Explanation:

Lo x 2 Lo x 3 Lo, Lo = 0.2 m,  K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,

time, t = 6 s

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.

What is the magnitude of the velocity after it hits the ground?

Answers

Answer:

9.25 m/s

Explanation:

A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps

Answers

Answer:

7.84 m/s

Explanation:

Height, h = 2 m

Initial velocity, u = 10 m/s

Angle, A = 80°

(a) Let the time taken to go to the net is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]

Time cannot be negative.

So, t = 0.2 s

The vertical velocity at t = 0.2 s is

v = u + at

v = 10 sin 80 - 9.8 x0.2

v = 9.8 - 1.96 = 7.84 m/s

A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?

Answers

Answer:

0.013 m/min

Explanation:

Applying,

dV/dt = (dh/dt)(dV/dh)............. Equation 1

Where

V = πr²h................ Equation 2

Where V = volume of the tank, r = radius, h = height.

dV/dh = πr²............ Equation 3

Substitute equation 3 into equation 1

dV/dt = πr²(dh/dt)

From the question,

Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14

Substitute these values into equation 3

2 = (3.14)(7²)(dh/dt)

dh/dt = 2/(3.14×7²)

dh/dt = 0.013 m/min

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.

What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.

Answers

Answer:

sup qwertyasdfghjk

Explanation:

1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.​

Answers

Answer:

r = 20.22 m

Explanation:

Given that,

Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]

Electric field, [tex]E=55\ N/C[/tex]

We need to find the distance. We know that, the electric field a distance r is as follows :

[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]

So, the required distance is 20.22 m.

The relation of mass m, angular velocity o and radius of the circular path r of an object with the centripetal force is-
a. F = m²wr
b. F = mwr²
c. F = mw²r
d. F = mwr. ​

Answers

Answer:

Correct option not indicated

Explanation:

There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).

The formula to calculate a centripetal force (F) is

F = mv²/r

Where m is mass, v is velocity and r is radius

where

While the formula to calculate a centrifugal force (F) is

F = mω²r

where m is mass, ω is angular velocity and r is radius of the circular path.

From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be C.

NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info

Answers

Answer:

The correct option is (E).

Explanation:

Given that,

Mass of object 1, m₁ = 1 kg

Mass of object 2, m₂ = 2 kg

They collides after the collision. We need to find the speed of the two boxes after the collision.

The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.

So, the information is not enough.

In 1.0 second, a battery charger moves 0.50 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery.
Part A:
How much work does the charger do? Answer is 0.75 J
Part B:
What is the power output of the charger in watts?

Answers

Answer:

W = Q * V     work done on charge Q

A. W = .5 C * 1.5 V = .75 Joules

B. P = W / t = .75 J / 1 sec = .75 Watts

If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.

Answers

Answer:

B. Smaller than

Explanation:

This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.

A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?

Answers

Answer:

Power = Energy/time

Energy = Power xtime.

Time= 20hrs

Power = 100Watt =0.1Kw

Energy = 0.1 x 20 = 2Kwhr.

This Answer is in Kilowatt-hour ...

If the one given to you is in Joules

You'd have to Change your time to seconds

Then Multiply it by the power of 100Watts.

a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss

Answers

Answer:

Explanation:

Apply law of conservation of momentum along y-axis.

Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.

Let the mass of each piece after breaking be m .

Momentum of piece moving along positive y-axis

= m x 10 = 10m .

Let the component of velocity of third piece along y-axis be v .

Its momentum along the same direction = m v .

Total momentum along y -axis = 10 m + m v

According to law of conservation of momentum

10 m + mv = 0

v = - 10 m/s .

Component of velocity of the third piece along y-axis will be - 10 m/s .

In other words it will be along negative y-axis with speed of 10 m/s.

You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abort the landing, so he pulls up, moving in a semicircular upward-bending path. The path has a radius of 450 m with a radial acceleration of 17 m/s^2.

Required:
What is the plane's speed?

Answers

Answer:

v = 87.46 m/s

Explanation:

The radial acceleration is the centripetal acceleration, whose formula is given as:

[tex]a_c = \frac{v^2}{r}[/tex]

where,

[tex]a_c[/tex] = centripetal acceleration = 17 m/s²

v = planes's speed = ?

r = radius of path = 450 m

Therefore,

[tex]17\ m/s^2 = \frac{v^2}{450\ m}\\\\v^2 = (17\ m/s^2)(450\ m)\\\\v = \sqrt{7650\ m^2/s^2}[/tex]

v = 87.46 m/s

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request

Answers

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

d. On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg>m3 . If the tank was full before the accident, what was the total outward force the molasses exerted on its sides

Answers

Answer:

F = 1.638 x 10⁸ N = 163.8 MN

Explanation:

The total force exerted by the molasses is given as:

F = PA

where,

F = Force exerted by the molasses = ?

P = Pressure = ρgh

ρ = density of molasses = 1600 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h = height of tank = 17.7 m

A = cross-sectional area of tank = πr²

r = radius of tank = 27.4 m/2 = 13.7 m

Therefore,

[tex]F = \rho ghA = \rho gh(\pi r^2)\\\\F = (1600\ kg/m^3)(9.81\ m/s^2)(17.7\ m)(\pi)(13.7\ m)^2[/tex]

F = 1.638 x 10⁸ N = 163.8 MN

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?

Answers

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

Help me with my physics, please

Answers

The right answer would be

-20t+ 80


Question 8 a-e plz

Answers

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??​

Answers

Answer:

Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)

prove mathematically :
1. v = u + at
2. s = ut+1*2 at ​

Answers

Answer:

a.v=u+v/2

a.v=s/t

combining two equation we get,

u+v/2=s/t

(u+v)t/2=s

(u+v)t/2=s

{u+(u+at)}t/2=s

(u+u+at)t/2=s

(2u+at)t/2=s

2ut+at^2/2=s

2ut/2+at^2/2=s

UT +1/2at^2=s

proved

a=v-u/t

at=v-u

u+at=v

A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction

Answers

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)

[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]f_{s}[/tex] - Static friction force, in newtons.

[tex]f_{k}[/tex] - Kinetic friction force, in newtons.

[tex]m[/tex] - Mass, in kilograms.

[tex]g[/tex] - Gravitational constant, in meters per square second.

If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:

[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{s} = 0.273[/tex]

[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.181[/tex]

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

two resistors with resistance values 4.5 ohms and 2.3 ohms are connected in series or parallel across a potential difference of 30V to a light bulb find the current flowing through the light bulb in both cases​

Answers

Answer:

Look at work

Explanation:

Series:

I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.

Parallel:

V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.

You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet

Answers

6 cubic feet I’m pretty sure that’s the answer

During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.

Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.

Answers

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

           

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

Answers

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

A block slides down a frictionless plane that makes an angle of 24.0° with the horizontal. What is the
acceleration of the block?

Answers

Answer:

F = m g sin theta      force accelerating block

m a = m g sin theta

a = 9.8 sin 24 = 3.99 m/sec^2

how do you calculate voltage drop

Answers

Answer:

Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.

Divide by 100.

Multiply by proper voltage drop value in tables. The result is voltage drop.

Explanation:

Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces

Answers

Answer:

The answer is C. Isolated System

Answer:

C. Isolated system

Explanation :

∵According to law of  conservation of momentum ,In an isolated system ,the total momentum remains conserved.

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