Answer:
[tex]v^2 = v_{ox}^2 + 2a(x - o_x)[/tex]
Explanation:
The options are not well presented; However, the questions can still be solved
Given
[tex]Initial\ Velocity= V_{ox}[/tex]
[tex]Initial\ Displacement = X_{o}[/tex]
[tex]Time = t[/tex]
Required
Determine the final displacement
This question will be answered using the following equation of motion
[tex]v^2 = u^2 + 2as[/tex]
Where s represent the total distance
s is the distance between the initial and final displacements and is calculated as thus;
[tex]s = x - o_x[/tex]
Where x represents the final displacement
Substitute [tex]V_{ox}[/tex] for [tex]u[/tex] ---- The initial velocity
[tex]v^2 = v_{ox}^2 + 2as[/tex]
Substitute [tex]x - o_x[/tex] for s
[tex]v^2 = v_{ox}^2 + 2a(x - o_x)[/tex]
Hence, the above equation can be used to determine the final displacement by solving for x
Find an expression for the square of the orbital period.
Answer:
T²= 4π²R³/GM
Explanation:
First we know that
Fg= Fc
Because centripetal force must equal gravitational force
So
GMm/R² = Mv²/R
But velocity is 2πR/T
So by substitution we have
GMm/R²= M (2πR/T)/T
We have
T²= 4π²R³/GM as period
What is the definition of Unbalanced Forces in Science ?
Two planets, Dean and Sam, orbit the Sun. They each have with circular orbits, but orbit at different distances from the Sun. Dean orbits at a greater average distance than Sam. According to Kepler's Third Law, which planet will have a longer orbital period? Group of answer choices Dean Sam Since they both have circular orbits, they will have the same orbital periods. There isn't enough information to tell.
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
Answer:
According to the law of universal gravitation, any two objects are attracted to each other. The strength of the gravitational force depends on the masses of the objects and their distance from each other.
Many stars have planets around them. If there were no gravity attracting a planet to its star, the planet's motion would carry it away from the star. However, when this motion is balanced by the gravitational attraction to the star, the planet orbits the star.
Two solar systems each have a planet the same distance from the star. The planets have the same mass, but Planet A orbits a more massive star than Planet B.
Which of the following statements is true about the planets?
A.
Planet B will keep orbiting its star longer than Planet A.
B.
Planet A has a longer year than Planet B.
C.
Planet A orbits its star faster than Planet B.
D.
Planet B is more attracted to its star than Planet A.
Explanation:
A solid, insulating sphere of radius 40.0 cm has positive charge distributed uniformly throughout its volume. The electric field at a distance of 80.0 cm from the center of the sphere is 6.00 N/C. What is the electric field at a distance of 20.0 cm from the center of the sphere (in N/C)
Answer:
The correct answer to the following question will be "12.0 N/C".
Explanation:
As we know,
Charged from the inside of the sphere throughout consideration of the electrical field or inside sphere.
⇒ [tex]E_0=\frac{KQ}{r_0^{2}}[/tex]
Now,
⇒ [tex]Q=\frac{E_0r_0^{2}}{k}[/tex]
On putting the values in the above formula, we get
⇒ [tex]=\frac{6.00\times 80.0}{9\times 10^9}(\frac{10^{-2}}{2})^2[/tex]
⇒ [tex]=4.267\times 10^{-10} \ C[/tex]
Electric field within the sphere at that same distance of 20.0 cm from either the core.
⇒ [tex]E_i=\frac{kQr_i}{R^3}[/tex]
On putting the values, we get
⇒ [tex]=\frac{(9\times 10^9)(4.267\times 10^{-10})(20.0)}{(80.0)^3(\frac{10^{-2}}{1} )^2}[/tex]
⇒ [tex]=12.0 \ N/C[/tex]
why is science important for our understanding of the natural world?
Which of these statements matches what we saw? Select all that apply: The amount of electric force depends on the amount of charge. Adding more charge to either balloon caused a greater repulsive force. If the one balloon has more electric charge, that causes a greater electric force on only on the other balloon. If the one balloon has more electric charge, that causes a greater electric force on only on itself, causing that balloon to move farther away than the other balloon. If the one balloon has more electric charge, that causes a greater electric force on both balloons, causing them to deflect equally.
Answer:
-The amount of electric force depends on the amount of charge
-Adding more charge to either balloon caused a greater repulsive force.
-If the one balloon has more electric charge, that causes a greater electric force on both balloons, causing them to deflect equally.
Explanation:
According to Coulomb's law of electrostatic force between two charged masses, the force of attraction is proportional to the product of the charges on the masses. This means that the more the charge on the masses, the more the electric force on them. And adding more charge to either balloon will only cause the balloons to repel each other with a greater amount of force.
The electric force between two charged masses is of the same magnitude both ways, regardless of the individual charges on the masses. This justifies the fact that If one balloon has more electric charge, it will cause a greater electric force to be exerted between both balloons, causing them to deflect equally.
Which figure shows vector A - vector B?
Answer:
The first and the last answer
While on vacation in the tropical Bahamas, Suzy decides to try out a water jet pack. She places the jet pack on her back (like you would wear a backpack), pushes the buttons in her hands, and water comes shooting out of the bottom of the jetpack, exerting a force on the ocean below. Why does Suzy fly up into the air? In your answer, be sure to state which of Newton's laws of motion this pertains to, describe this law, and explain how it causes Suzy to fly up into the sky.
Answer:
-Suzy flies up into the sky because of the upward force on the jetpack.
-Newton's third law of motion can be ascribed to this.
Explanation:
We know that in momentum, Force is described as the rate of change of momentum. Thus, for us to understand the forces, we will need to establish how the momentum of the water changes.
Suppose the mass flow rate of the water is denoted as M kg/sec and the upward velocity is denoted as v_u.
Now, since water is lying stationary at the sea, the initial momentum of the water will be zero. If every second the boat pumps upwards with mass flow rate M at velocity vu, the momentum per second would be;
Δp = F_water = Mv_u
Now, from Newton's third law the boat/pump would experience a downwards force with the formula:
F_boat = −Mv_u
Furthermore at the jetpack, the water does a reverse turn in direction and is now again directed back down at a different velocity v_d. Therefore the final momentum per second is Mv_d and therefore change in momentum is now;
Δp = M(v_d - v_u)
The velocity v_d will be negative since the boat/pump would experience a downwards force. Thus, the force on the water will be downwards and F_water ≤ 0 .
We will finally use Newton's third law to conclude that the upward force on the jetpack will be;
F_jetpack = M(v_u - v_d)
Thus upward force is what makes Suzy to fly up in the sky.
An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?
Answer:
a) Magnitude = 1.03 m/s², Direction: south
b) [tex]V_{f} = 8.16 m/s [/tex]
Explanation:
a) The magnitude and direction of the acceleration can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex] (1)
Where:
[tex]V_{f}[/tex]: is the final speed = 9.40 m/s
[tex]V_{0}[/tex]: is the initial speed = 13.0 m/s
t: is the time = 3.50 s
Solving equation (1) for a, we have:
[tex] a = \frac{V_{f} - V_{0}}{t} = \frac{9.40 m/s - 13.0 m/s}{3.50 s} = -1.03 m/s^{2} [/tex]
Hence, the magnitude of the acceleration is 1.03 m/s² and the direction of the bird's acceleration is the opposite of the initial velocity direction, which means that the bird is decelerating.
b) The final velocity of the bird can be found using the same equation 1:
[tex] V_{f} = V_{0} + at [/tex]
[tex] V_{f} = 13.0 m/s + (-1.03 m/s^{2})*(3.50 s + 1.20 s) = 8.16 m/s [/tex]
Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.
I hope it helps you!
10.A car is travelling at a constant speed of 27m/s. The driver looks away from the road for a 2.0s to tune in a station on the radio. How far does the car go during this time?
Explanation:
Distance = speed × time
d = (27 m/s) (2.0 s)
d = 54 m
b. How tar from the li u
b. Obtain the value of g from the motion of the moon assuming that its period
of rotation round the earth is 27 days 8 hours and the radius of its orbit is
60.1 times the radius of the earth.
0.
7.2 m. Compute
Answer:
Approximately [tex]9.79\; \rm m \cdot s^{-2}[/tex] at the surface of the earth, given that the radius of the earth is known to be approximately [tex]6.371\times 10^{6}\; \rm m[/tex]. Assumptions: the earth is a sphere, the orbit of the moon is circular, and that the gravitational pull of the earth is the only force on the moon.
Explanation:
Convert the orbital period of the moon around the earth to seconds:
[tex]\begin{aligned}t &= (27\; \text{day} \times 24\; \text{hours} \cdot \text{day}^{-1} + 8\; \text{hour})\times 3600\; \rm \text{s}\cdot \text{hour}^{-1}\\ &= 2361600\; \text{s} \end{aligned}[/tex].
Calculate the angular velocity of the moon around the earth using the formula[tex]\displaystyle \text{angular velocity} = \frac{2\pi}{\text{orbital period}}[/tex]:
[tex]\begin{aligned}\omega = \frac{2\pi }{t} = \frac{2\pi}{2361600\; \rm s} \approx 2.66056\times 10^{-6}\; \rm s^{-1}\end{aligned}[/tex].
Let [tex]m(\text{moon})[/tex] denote the mass of the moon. Let [tex]g[/tex] denote the gravitational field strength at the surface of the earth (not at the position of the moon.) Let [tex]r[/tex] denote the radius of the earth.
The orbital radius of the moon would thus be [tex]60.1\, r[/tex]. Note that in the gravitational field due to a single spherical mass, the field strength is inversely proportional to the square of distance from the center of that mass.
The surface of the earth is at a distance of [tex]r[/tex] away from the center of the earth. The distance between the moon and the center of the earth would then be [tex]60.1[/tex] times that number (that is: [tex]60.1\, r[/tex].)The gravitational field at the surface of the earth is [tex]g[/tex]. Therefore, at the position of the moon ([tex]60.1[/tex] times further away from the center of the earth compared to the earth surface,) the gravitational field strength would be:
[tex]\displaystyle \left( \frac{r^2}{(60.1\, r)^2} \right)\, g = \frac{g}{60.1^2}[/tex],
Hence, the gravitational pull of the earth on the moon would be [tex]\displaystyle \frac{m(\text{moon})\, g}{60.1^2}[/tex].
Assume that the gravitational pull of the earth is the only force on the moon. That gravitational pull would then be the equal (in size) to the net force on the moon. That is:
[tex]\displaystyle \text{Net force on the moon} = \frac{m(\text{moon})\, g}{60.1^2}[/tex].
On the other hand, the rotation (assumed to be perfectly circular) of the moon would give the net force on the moon in terms of:
the mass of the moon, the angular speed of the rotation, andthe radius of the orbit.[tex]\displaystyle \text{Net force on the moon} = m(\text{moon})\, \omega^2\, (60.1\, r)[/tex].
Combine the two expressions for the net force on the moon to obtain an equation for [tex]g[/tex]:
[tex]\displaystyle \frac{m(\text{moon})\, g}{60.1^2} = m(\text{moon})\, \omega^2\, (60.1\, r)[/tex].
Simplify and solve for [tex]g[/tex]:
[tex]\displaystyle g = 60.1^3 \, \omega^2\, r[/tex].
The angular velocity of this rotation, [tex]\omega[/tex], has already been found to be approximately [tex]2.66056\times 10^{-6}\; \rm s^{-1}[/tex]. Look up the radius of the earth: [tex]r = 6.371\times 10^{6}\; \rm m[/tex]. Evaluate this expression for [tex]g[/tex]:
[tex]\begin{aligned}g &= 60.1^3 \, \omega^2\, r \\ &\approx 60.1^3 \times \left( 2.66056\times 10^{-6}\; \rm s^{-1}\right) \times 6.371\times 10^{6}\;\rm m \\ &\approx 9.79\; \rm m \cdot s^{-1}\end{aligned}[/tex].
Philip drives his car at a velocity of 28 m/s he applies the break which slows the vehicle down at a rate of six. 4 m/s2 and it causes it’s to slow to stop how long does it take for the car to stop answer to the nearest 10th?
A Me and Ed's Pizza has 16 slices. John eat 7/12pieces of pizza. Harold
finished 4/23 pieces of pizza. If there was only 23of a piece left over, how
much pizza did Betty eat?
Answer:
Explanation:
Total number of slices in the pizza = 16
Total number ate by John,
J = 7/12 * 16 = 28/3 = 9.3
Total number ate by Harold,
H = 4/23 * 16 = 2.783
Total number left over
L = 1/23 * 16 = 0.696
Total number ate by Betty
B = ?
All we have to do to get the total number eaten by Betty is to sum up that eaten by Harold and John, including the left over, then subtract it from the total pizza slices.
Betty = Total - (Harold + John + Left Over)
Betty = 16 - (9.3 + 2.783 + 0.696)
Betty = 16 - 12.779
Betty = 3.221
Therefore, Betty ate 3.221 of the total pizza as a whole
At which point in time does an object with the motion represented in the graph have an
instantaneous acceleration of -2 m/s^2?
Answer:
4s.
Explanation:
Hello,
In this case, considering that the instantaneous acceleration is computed via the change in the velocity divided by the change in the time, at 4 s we can evidence that the instantaneous acceleration is -2m/s² since the initial velocity at t=0s is 8m/s and at t=4s the velocity is 0 m/s, meaning that such decrease accounts for a deceleration process which is represented by a negative acceleration, and the value is:
[tex]a=\frac{0m/s-8m/s}{4s-0s}\\ \\a=-2m/s^2[/tex]
Best regards.
The flow rate of water through a tapered straight horizontal pipe fitting is 5 m/s. The diameter at the entrance is 0.7 m and is reduced by a factor of 0.6 at the exit. If the gauge pressure at the entrance is 350 kPa and drops by 50 Pa, the horizontal thrust on the fitting is,
Answer:
The value is [tex]F_t = 76024 \ N[/tex]
Explanation:
From the question we are told that
The flow rate is [tex]v_1 = 5 \ m/s[/tex]
The entrance diameter is [tex]d = 0.7 \ m[/tex]
The exit diameter is evaluated as [tex]d_e = 0.6 * 0.7 = 0.42 \ m[/tex]
The entrance gauge pressure is [tex]P_g = 350 \ kPa = 350*10^{3} \ Pa[/tex]
The droped gauge pressure is [tex]P_d = 50 \ kPa = 50*10^{3} \ Pa[/tex]
The presure at the exist is evaluated as [tex]P_e =(350 - 50 ) \ kPa = 300*10^{3} \ Pa[/tex]
Generally the entrance cross-sectional area is mathematically represented as
[tex]A = \pi \frac{d^2}{4}[/tex]
[tex]A = 3.142 \frac{0.7^2}{4}[/tex]
[tex]A =0.385 \ m^2[/tex]
Generally the exit cross-sectional area is mathematically represented as
[tex]A_e = \pi \frac{d_e^2}{4}[/tex]
[tex]A_e = 3.142 \frac{0.42^2}{4}[/tex]
[tex]A_e =0.139\ m^2[/tex]
Generally from the continuity equation
[tex]v_1 * A = v_2 * A_e[/tex]
=> [tex]v_2 = \frac{v_1 * A}{A_e}[/tex]
=> [tex]v_2 = \frac{5 * 0.385}{0.139}[/tex]
=> [tex]v_2 = 13.8 m/s [/tex]
Generally the net force in horizontal axis is equivalent to the net momentum change in the horizontal direction
So
[tex]P_g * A + F_t - P_e * A_e = P_d [ A_e * v_2^2 - A* v_1^2][/tex]
Here [tex]F_t[/tex] is the horizontal thrust on the fitting
So
[tex]350*10^{3} * 0.385 + F_t -0.385 * 0.139 = 50*10^{3} [ 0.385* 13.8^2 - 0.385* 5^2][/tex]
An average human has a heart rate of 70 beats per minute. If someone's heart
beats at that average rate over a 70-yr lifetime, how many times would it beat?
a.) 7 x 10 to the 5th power
b.) 2 x 10 to the 6th power
c.) 2x 10 to the 7th power
d.) 3 x 10 to the 9th power
A fully loaded elevator at maximum capacity weighs 2400 lbs. The counterweight weighs 1000 lbs. The elevator always starts from rest at its maximum acceleration of g/4 whether it is going up or down. (a) What force does the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up
Answer:
2250lb
Explanation:
Using
1000-T=(1000/g)(g/4)
So
T = 750lb
Then
750+F-2400= 2400/9
So F= 2250lb
Car B is following car A as they are moving along a straight path with vA=40 mph and vB=45 mph. At the moment when the distance between the cars is 45 ft brakes are applied simultaneously in both cars. Car A decelerates with aA=−22 ft/s2 and car B with aB=−20 ft/s2. What is the distance between the cars when they are both stopped?
Answer:
s = 14.3 ft
Explanation:
First we need to calculate the distances traveled by both the cars. We use third equation of motion for that:
2as = Vf² - Vi²
where,
a = acceleration
s = distance
Vf = Final Velocity
Vi = Initial velocity
FOR CAR A:
Vi = Va = (40 mph)(5280 ft/1 mile)(1 h/3600 s) = 58.66 ft/s
Vf = 0 ft/s
a = aA = - 22 ft/s²
s = sa = ?
Therefore,
2(- 22 ft/s²)(sa) = (58.66 ft/s)² - (0 ft/s)²
sa = 78.2 ft
FOR CAR B:
Vi = Vb = (45 mph)(5280 ft/1 mile)(1 h/3600 s) = 66 ft/s
Vf = 0 ft/s
a = aB = - 20 ft/s²
s = sb = ?
Therefore,
2(- 20 ft/s²)(sb) = (66 ft/s)² - (0 ft/s)²
sb = 108.9 ft
Since, the car A was initially 45 ft ahead of car B. Therefore,
sa = 45 ft + 78.2 ft = 123.2 ft
Now, the distance between the cars will be:
s = sa - sb
s = 123.2 ft - 108.9 ft
s = 14.3 ft
Two ropes are attached to a tree, and forces of F⃗ 1=2.0iˆ+4.0jˆN and F⃗ 2=3.0iˆ+6.0jˆN are applied. The forces are coplanar (in the same plane). (a) What is the resultant (net force) of these two force vectors? (b) Find the magnitude and direction of this net force.
Answer:
We can see that the 2 forces are being applied in the same direction
So the resultant force will be larger than the given forces
Resultant force = (2i + 4j) + (3i + 6j)
R = 5i + 10j
Magnitude of the resultant force :
R² = i² + j²
R² = 25 + 100
R = [tex]\sqrt{125}[/tex] = [tex]5\sqrt{5}[/tex]
Direction of the resultant force:
Tan Θ = Vertical component of force / Horizontal component of force
Tan Θ = 10 / 5
Tan Θ = 2
Θ = Arctan (2)
Θ = 63.4 degrees
Direction is 63.4 degrees in the NE direction
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A) The resultant ( net force ) = 5i + 10 j
B) The magnitude and direction of this net force
magnitude of net force = 5√5 direction of net force = 63.4° NEGiven that :
The vector forces are coplanar ( In the same plane and direction )
A) resultant force = (2i + 4j) + (3i + 6j) = ( 2 + 3 ) i + ( 4 + 6 ) j
= ( 5 i + 10 j )
B ) Calculate The magnitude and direction of the net force
i) Magnitude of the force
R² = ( i² + j² )
= ( 5² + 10² )
= 25 + 100
∴ R = √ (25 + 100) = 5√5
ii) Determine the direction of resultant force
Tan ∅ = opposite / adjacent ( vertical force ( j ) / horizontal force ( i ) )
= 10 / 5 = 2
∅ = arctan ( 2 ) ≈ 63.4°
∴ direction of resultant force = 163.4° NE
Hence we can conclude that the resultant ( net force ) = 5i + 10 j and The magnitude and direction of this net force
magnitude of net force = 15 direction of net force = 1.107° NELearn more : https://brainly.com/question/24034108
Counseling psychologists typically handle severe psychological disorders.
Please select the best answer from the choices provided
T
F
Answer:
FALSE
Explanation:
Answer:
False
Explanation:
When an individual registers sensory input without being aware of the process consciously, __________ occurs. subliminal stimulation sensory adaptation signal detection processing absolute threshold
Answer:
Subliminal simulation
Explanation:
When an individual registers sensory input without being aware of the process consciously, Subliminal simulation occurs.
What is Subliminal simulation?Subliminal stimulation defined as sensory stimulation which is below a person's threshold for perception.
It can't be seen by the bare eye or consciously heard. An example is visual stimuli which is flashed so quickly on a screen that a person can't process it. So, they are unaware they have seen anything.
Subliminal stimuli helps facilitate conscious processing of related information, changes the current mood, boost our motivation. It can even change a person's political attitudes and voting intentions.
When an individual registers sensory input without being aware of the process consciously, Subliminal simulation occurs.
Learn more about Subliminal simulation
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Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? Atction force:_______ Reaction Force:________
Answer: action forc roketorce
reaction force is engine fires
in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and move together. What is the total momentum after the collision? 55 kg*m/s 275 kg * m/s -55kg * m/s -275 kg m/2
Explanation:
The momentum of the three objects are as follow :
11 kg-m/s, -65 kg-m/s and -100 kg-m/s
Before collision, the momentum of the system is :
[tex]P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s[/tex]
After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.
It would mean that, after collision, momentum of the system is equal to the initial momentum.
Hence, final momentum = -154 kg-m/s.
URGENT HELP!! PLEASE ANSWER QUICK
Answer:
im pretty sure it is B or D
A car is going 70 MPH down the road and hits a tree. The passenger is not wearing a seatbelt and goes through the windshield, explain why the driver remained in the car and the passenger did not. Be sure reference Newton’s first law in your explanation
Answer:
Inertia is the reason that people in cars need to wear seat belts. Instead, the riders continue moving forward with most of their original speed because of their inertia. If the driver is wearing a seat belt, the seat belt rather than the windshield applies the unbalanced force that stops the driver's forward motion.
Answer:
Explanation:
The driver was able to stay in the car because he or she had a seat belt on, the passenger did not have a seat belt on, therefore they were not able to stay in the car. Newton’s first law states “An object in motion stays in motion unless acted on by another force”. Therefore the passenger was yoinked out of the car.
8. An object that is traveling 50m/s slows down with an
acceleration rate of -3.5m/s/s. This took place over a period of 4s.
What was the objects final velocity? *
Answer:
36 m/s
Explanation:
Given:
v₀ = 50 m/s
a = -3.5 m/s²
t = 4 s
Find: v
v = at + v₀
v = (-3.5 m/s²) (4 s) + 50 m/s
v = 36 m/s
Light is incident along the normal on face AB of a glass prismwith refractive index 1.52.
A. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in air.
B. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in water.
air refactive index = 1.00029
water refractive index = 1.33
Answer:
48.9°, 29°
Explanation:
We kno that snell's law States that
( sin i / sin r ) = ( n 2 / n 1 )
And here
i = 90 - α
r = 90
n 2 = n a = 1.00029
n 1 = n = 1.52
So
( sin ( 90 - α ) / sin 90 ) = (1.00029 / 1.52 )
= 0.658
sin ( 90 - α ) = 0.658
90 - α = sin -1 ( 0.658 )
= 41.15
α = 90 - 41.15
= 48.9°
( b ) to find angle when prism immerssed in water
Using snell's law again
( sin i / sin r ) = ( n 2 / n 1 )
here i = 90 - α
r = 90
n 2 = n w = 1.33
n 1 = n = 1.52
So
( sin ( 90 - α ) / sin 90 ) = (1.33 / 1.52 )
= 0.875
sin ( 90 - α ) = 0.875
90 - α = sin -1 ( 0.875 )
= 61
α = 90 - 61
= 29°
An elephant pushes with 200 N on a load of trees. it then pushes these trees for 10 N. How much work did the elephant do?
Answer:
the answer is 2000Nm
Explanation:
wprk done = force × distance moved
w.d = 200N × 10m
w.d = 2000Nm
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Law of conservation of energy states that
a
Energy cannot be created or destroyed
b
Momentum is not lost, it is transferred
c
The sum of the kinetic energy and potential energy
You get out of school and walk north for 20 minutes to travel 1.6 km what is your average velocity?
Answer: 12.5 km/h
Explanation: 20/1.6 = 12.5 km/h
The average velocity of an individual will be "0.08 km/min". To understand the calculation, check below.
Distance and VelocityAccording to the question,
Time, t = 20 minutes
Distance, d = 1.6 km
We know the formula,
→ Distance = Speed × Time,
Or,
Speed, v = [tex]\frac{Distance}{Time}[/tex]
By substituting the values,
= [tex]\frac{1.6}{20}[/tex]
= 0.08 km/min
Thus the avg. velocity will be "0.08 km.min" is correct.
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