Answer:
B
Explanation:
The centripetal force keeps an object moving in a circular path. Therefore option (B) is correct.
What is centripetal force?A centripetal force can be described as a force that makes a body follow a curved path and its direction is orthogonal to the motion of the body. Gravity offers the centripetal force causing astronomical orbits.
The centripetal force is directed perpendicular to the direction of the displacement of an object. It always acts towards the center of the circle on an object moving in a circular path. For example, When spinning a ball on a string, the tension on the rope pulls the object toward the center.
The Centripetal Force can be described as the product of mass and velocity squared, divided by the radius.
F = mv²/r
Where F is the Centripetal force, m is the mass, r is the radius of the circle and v is the velocity of the object.
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(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answer:
a = 72 m/s^2
Explanation:
Since the object is moving in circle, the acceleration points TOWARDS the centre of the circle (it is centripetal acceleration)
a = v^2 / r
a = 60^2 / 50
a = 72 m/s^2
if mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion
Answer:
See explanations below
Explanation:
According to Newtons second law of motion
F = mass * acceleration
F = ma
If mass of an object is decreased to half, then m₂ = 1/2 m
If acting force is reduced by quarter, then F₂ = 3/4 F
F₂ = m₂a₂
3/4F = 1/2m a₂
Divide both expressions
(3/4F)/F = (1/2m)a₂/ma
3/4 = 1/2a₂/a
3/4 = a₂/2a
4a₂ = 6a
2a₂ = 3a
a₂ = 3/2 a
Hence the acceleration of its motion will be one and a half of its original acceleration.
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
Find all of those elements on the periodic table
Answer:
hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium
Explanation:
Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
A train with proper length L has clocks at the front and back. A photon is fired from the front to the back. Working in the train frame, we can easily say that if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c. Now consider this setup in the ground frame, where the train travels by at speed v. Rederive the above frame-independent result (namely, if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c) by working only in the ground frame.
Explanation:
In train's rest frame, the speed of photon is [tex]c[/tex] and the proper length of the train is [tex]L[/tex]. The time taken by the photon to cross the train is [tex]t=\frac{L}{c}[/tex]
In ground frame, the speed of the photon is given as follows:
[tex]v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}[/tex]
[tex]=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c[/tex]
The speed of light or photon remains same in every frame of reference.
Now, the speed of train is very less as compared to the speed of photon so that [tex]v<c[/tex] So that, [tex]\frac{v}{c} \ll 1[/tex]
The length contraction in the ground frame is given as follows:
[tex]L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}[/tex]
[tex]=L[/tex]
Time taken by the photon to travel the length of the train in ground frame is .
What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)
total lunar eclipse
total solar eclipse
partial lunar eclipse
partial solar eclipse
Answer: I'm not sure, but I think it would be a total lunar eclipse
When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.
What is partial lunar eclipse?A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.
During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.
A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.
The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.
Thus, the correct option is C.
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३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
DUE TODAY BEST ANSWER GET BRAINLIE
Which statements explain the relationship between mass, velocity and kinetic energy? * Select the TWO (2) that apply.*
Consider the equation to calculate kinetic energy: KE = ½mv2
Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.
Kinetic energy decreases if either the mass or the velocity of the object increases or if both increase.
Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease.
Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.
Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.
Explanation:
These statements explain the relationship between mass, velocity, and kinetic energy .i.e KE = ½mv², (3) Kinetic energy decreases if either the mass or the velocity of the object decreases or if b decrease. (4)Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.
What is kinetic energy?When a body has a mass of m kg and travels a velocity of v m/s then the kinetic energy is given by,
KE=1/2mv²
In the above relation, it is clearly shown that,
Kinetic energy is directly proportional to the mass of the body,
i.e
KE∝m
and also Kinetic energy directly varies with the square of the velocity of the body.
I.e
KE∝v²
Therefore, From the given expression we can conclude that when mass and velocity increased either one of them or both, the kinetic energy also gets increased.
And when mass and velocity decreased either one of them or both, the kinetic energy also gets decreased. Hence options (3) and (4) are correct.
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Even in the most advanced circuits, we cannot oscillate electrons back and forth at that rate through wires. But we can oscillate charges back and forth quickly enough to broadcast TV using radio wave signals. At what frequency do the electronics at the TV station need to have the charges oscillate back and forth on a TV broadcast antenna to transmit a typical TV signal (say a radio wave transmission signal with a wavelength of 1 meter)
Answer:
the oscillations of the electrons must be in the 10⁸ Hz = 100 MHz range
Explanation:
The speed of a wave of radio, television, light, heat, all are manifestations of electromagnetic waves that are oscillations of electric and magnetic fields that support each other, the speed of all these waves is the same and the vacuum is equal to c = 3 108 m / s
All waves have a relationship between the speed of the wave, its frequency and wavelength
c = λ f
f = c /λ
for this case lam = 1 m
f = 3 10⁸/1
f = 3 10⁸ Hz
the oscillations of the electrons must be in the MHz range
It should be clarified that the speed of light in air is a little lower
n = c / v
v = c / n
the refractive index of vacuum is n = 1 and the refractive index of air is n = 1.000002
discuss two reasons why people find transition between school and university
Answer:
Is that your answer
A strong electromagnet produces a uniform magnetic field of 1.60 T and a cross-sectional area of 0.200 m2. If we place a coil having 240 turns and a total resistance of 21.0 around the electromagnet and then we then smoothly reduced the current in the electromagnet until it reaches zero in 20.0 ms, what is the current induced in the coil
Answer:
the magnitude of the induced current is 182.86 A.
Explanation:
Given;
number of turns, N = 240 turns
cross sectional area of the loop, A = 0.2 m²
uniform magnetic field strength, B = 1.6 T
resistance of the loop, R = 21 ohms
time, Δt = 20.0 ms
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA B}{t} \\\\emf = \frac{240 \times 0.2 \times 1.6}{20 \times 10^{-3}} \\\\emf = 3,840\ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{3840}{21} \\\\I= 182.86 \ A[/tex]
Therefore, the magnitude of the induced current is 182.86 A.
How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius
Answer:
[tex]3.52\times 10^{25}\ \text{J}[/tex]
Explanation:
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]
m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]
[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]
Energy required is given by
[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]
The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].
A 50kg crate is being push on a horizontal floor at constant velocity. Given that the coefficient of kenitic friction between crate and floor is hk=0.1 . What is the push force?
Answer:
F = 49 N
Explanation:
For this exercise we must use Newton's second law. Let's set a reference frame with the x axis parallel to the floor.
As they indicate that the box is going at constant speed, its acceleration is zero
Y axis y
N-W = 0
N = mg
X axis
F-fr = 0
F = fr
the friction force has the formula
fr = μ N
fr = μ mg
we substitute
F = μ m g
let's calculate
F = 0.1 50 9.8
F = 49 N
How can Newton’s laws of motion and the law of gravitation predict the motion of an object, and how can frames of reference be used to describe that motion?
What factors affect the speed of a wave? Check all that apply.
the amplitude of the wave
the energy of the wave
the temperature of the medium
the type of wave
the type of medium
Answer:
the amplitude of the wave
the energy of the wave
the type of wave
the type of medium
A 40 kg rock is rolling toward a town at 4 m/s after an earthquake. Calculate the KE.
Be sure to show your work and include units!
Use the formula KE = 1/2mv2 will brainlist
Answer:
320 J
Explanation:
From the question,
KE = 1/2mv².................. Equation 1
Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock
Given: m = 40 kg, v = 4m/s
Substitute these values into equation 1
KE = 1/2(40)(4²)
KE = 20×16
KE = 320 J
Hence the kinetic energy of the rock is 320 J
A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the fist moves through an arc length of 100 cm. What is the average speed of the fist during the hook
Answer:
average speed of the fist during the hook = 15 m/s or 1500 cm/s
Explanation:
We are given;
Speed of shoulder contraction, v_s = 75 cm/s = 0.75 m/s
Distance moved through the arc length by shoulder, d_s = 5 cm = 0.05 m
Distance moved by the fist, d_f = 100 cm = 1 m
Now, we are to find the average speed of the fist during the hook; v_f
Thus can be gotten from proportion;.
d_f/d_s = v_f/ v_s
Making V_f the subject, we have;
v_f = (d_f × v_s)/d_s
Thus;
v_f = (1 × 0.75)/0.05
v_f = 0.75/0.05
v_f = 15 m/s
What type of fault is shown in the image below?
In this fault, the hanging wall and footwall are side by side and moving in opposite directions.
Public Domain
Lateral fault
Normal fault
Reverse fault
Thrust fault
Answer:
this is a lateral fault
Explanation:
Four children climb on a carousel that is initially at rest. If the carousel accelerates to 0.4 rad/s within 10 seconds, what is the angular acceleration? If the carousel has a radius of 20 meters, what is the linear velocity of the children?
Answer:
If the angular acceleration is a constant A, then this constant times the time will be equal to the angular velocity.
We know that after accelerating for 10 seconds, the angular velocity is 0.4 rad/s
Then:
A*10s = 0.4 rad/s
If we solve this for A we get:
A = (0.4 rad/s)/10s = 0.04 rad/s^2
This is the angular acceleartion.
Now we assume that the angular speed remains constant at 0.4 rad/s, and we know that the radius of the carousel is 20m, we want to find the linear velocity of the children.
We know that for an angular velocity W and for a radius R, the linear velocity is:
V = R*W
Then in this case the linear velocity is:
V = 20m*(0.4 s^(-1)) = 8 m/s
A large wooden block of weight 30n, is observed to push down a ramp. The block is pushed down an inclined plane as shown below and the applied force is parallel to the plane and has a magnitude of 5 n as shown. The force of kinetic friction is 17n the block is pushed 6m down along the inclination of the ramp is 30 degrees with the horizontal
Answer:
OMG IM ON THE SAME QUESTION
Explanation:
An object, accelerating from rest at a constant rate, travels over 28 m in 11 s. What is its final velocity?
1.81 m/s
3.20 m/s
5.09 m/s
0.00 m/s
none of the above
Answer:
answer is 3
Explanation:
by using s= [(v+ u)/2] x t
28= (v+ 0)/2 x 11
v= 5.09 ms^-1
1. Define force and give its Sl unit
Answer:
Force is an external agency that changes or tends to change the state of body from rest to motion or motion to rest.
The SI unit of force is newton(N)
a sharp image is formed when light reflects from a
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
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During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.145 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.5 m/s , shatters the glass as it passes through, and leaves the window at 10.9 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball
Answer:
J = -0.522 m/s
Explanation:
Given that,
The mass of the baseball, m = 0.145 kg
Initial velocity, u = 14.5 m/s
Final velocity, v = 10.9 m/s
aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,
[tex]J=m(v-u)[/tex]
Substitute all the values,
[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]
The direction of impulse is opposite to the direction of velocity.
Select four of the following that would increase the magnetic field of an electromagnet
Answer:
The correct answers are: A, C, D, E
Explanation:
The magnetic field is a solenoid is given by
B = μ₀ [tex]\frac{N}{L}[/tex] I
where N is the number of turns, I the current and L length of the solenoid.
Using this equation let's examine the different responses to permute increasing the magnetic field
A) True. a thicker wire decreases the resistance and the current can increase the system.
B) False. If there is no voltage source there is no current in the system
C) True. the field is proportional to the number of turns
D) True. the magnetic moments of the core align with the field increasing its value
E) True. When the loops are closer together, more of them can fit per unit length
F) False. If the wire is shorter the number of turns decreases.
The correct answers are: A, C, D, E
When we add or remove energy from a substance, what kind of changes can we observe? Can they happen at the same time ?
Answer:
When you add energy to a substance?
One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules. The temperature at which a substance goes from a solid to a liquid is it melting point.
When you remove energy from a substance?
When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.
Explanation:
The addition of energy increases the kinetic energy of the particles, which reduces the intermolecular forces between the particles. Freezing occurs when a liquid becomes a solid and energy is released.
Atoms lose energy when they change from solid to liquid or gas and quid to gas.
A camper hears her echo off a
hillside 1.52 s after yelling. How
far is she from the hill?
Answer:
261
Explanation:
Acellus
Which is an example of a noncontract force?
(A) elastic force
(B) Normal force
(C)Applied force
(D) Electric Force
Answer:
electric force
Explanation:
its a contact and noncontact force
A baseball with a mass of 142 g is thrown horizontally with a speed of 38.6 m/s (86 mi/h) at a bat. The ball is in contact with the bat for 1.00 ms and then travels straight back at a speed of 46.1 m/s (103 mi/h). Determine the average force (in N) exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)
Answer:
-11.600N
Explanation:
lol