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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A planet with a mass one-half that of Earth has a radius that is 3 times that of Earth's radius. What is the gravitational field strength of the planet?
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's (1/18) as much as on Earth.
The acceleration of gravity there would be about 0.545 m/s².
This is about 12% less than the gravity on Pluto.
What is the scientific study of how animals are classified?
Answer:
biology
Explanation:
Animals and plants
As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Answer:
102.5N
Explanation:
Given that a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W
The power = 4.12 × 746 = 3073.52 W
Using the formula
Power = force × velocity
3073.52 = force × 30
Force = 3073.52 / 30
Force = 102.5 N
Since most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road, therefore,
the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s is 102.5 N
prove that d1=R(d1-d2) in relative density
Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.
Answer:
[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]
[tex]44.43^{\circ}[/tex], second order does not exist
Explanation:
n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]
[tex]\lambda[/tex] = Wavelength
m = Order
Distance between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]
[tex]\lambda=400\ \text{nm}[/tex]
m = 1
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]
The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].
[tex]\lambda=700\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]
Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.
The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].
Second order angle does not exist.
If we always drop the balls from 1-m
height in each trial, what type of variable
is this in this experiment?
Constant Variable
Dependent Variable
Independent Variable
Answer:
height
weight of ball
time of ball falling
A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed
Answer:
The angular speed is 23.24 rad/s.
Explanation:
Given;
mass of the disk, m = 7 kg
radius of the disk, r = 0.2 m
applied force, F = 42 N
distance moved by disk, d = 0.9 m
The torque experienced by the disk is calculated as follows;
τ = F x d = I x α
where;
I is the moment of inertia of the disk = ¹/₂mr²
α is the angular acceleration
F x r = ¹/₂mr² x α
The angular acceleration is calculated as;
[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]
The angular speed is determined by applying the following kinematic equation;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]
initial angular speed, ωi = 0
angular distance, θ = d/r = 0.9/0.2 = 4.5 rad
[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]
Therefore, the angular speed is 23.24 rad/s.
While a boulder is on top of a hill, it has kinetic energy.
True
False
(5 Points)
a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)
Answer:
1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Explanation:
1. From Boyles' law;
[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]
[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa
[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]
[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]
Thus,
1 x [tex]10^{5}[/tex] x 9.6 = [tex]P_{2}[/tex] x 12
[tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]
= 80000
[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. Pressure, P = ρhg
where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).
Thus,
P = 1020 x 15 x 9.8
= 149940
P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
which energy conservation takes place when a toaster is switched on?
Answer:
A toaster usually takes in electrical energy.
Two types of energy are created.
In the first step, all of the input electrical energy is transformed into heat energy. That heat goes first into the coils or heating elements.
The heating elements get hot and glow.
The hot elements then transfer thermal energy (heat energy) into the air inside the toaster.
Explanation:
Answer: electrical energy
Explanation: An electric toaster takes in electrical energy from the power outlet and converts it into heat, very efficiently. If you want your toast to cook quickly, you need a toaster that radiates as much heat as possible each second onto your bread. hope this helps. Can u pls give me brainliest
A motorist is driving at 15 m/s when she sees that a traffic light 315m ahead has just turned red. She knows that this light stay red for 25 s, and she wants to be 20 m from the ligt when it turns green again. Taht way, she will still be able to stop if the light stays red longer than expected. She applies the brake gradually such that her acceleration is ax(t)= c + bt, where c and b are constant. Assume she starts with a constant speed at the origin.
Find the values of c a b and any other unknown constants in order to answer the following questions.
1. Given the motorist's acceleration as a function of time, what are her position and velocity fucntions? - Do not use numbers for any constant here. Only derive the position and velocity functions.
2. What is her speed as she reaches the light?
Answer:
1) x = x₀ + vot - ½ c t² - 1/6 bt³, v = v₀ - ct - ½ b t²
2) v₁ = 5.25 m/s, v₂ = -8 m/s
Explanation:
1) For this exercise, the relationship of the body is not constant, so you must use the definition of speed and position to find them.
acceleration is
a = c + bt
a) the relationship between velocity and acceleration
a = [tex]\frac{dv}{dt}[/tex]
dv = -a dt
The negative sign is because the acceleration is contrary to the speed to stop the vehicle.
we integrate
∫ dv = - ∫ a dt
∫ dv = -∫ (c + bt) dt
v = -c t - ½ b t²
This must be valued from the lower limit, the velocity is vo, up to the upper limit, the velocity is v for time t
v - v₀ = -c (t-0) - ½ b (t²-0)
v = v₀ - ct - ½ b t²
b) the velocity of the body is
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
we replace and integrate
∫ dx = ∫ (v₀ - c t - ½ bt²) dt
x-x₀ = v₀ t - ½ c t² - ½ b ⅓ t³
Evaluations from the lower limit the body is at x₀ for t = 0 and the upper limit the body is x = x for t = t
x - x₀ = v₀ (t-0) - ½ c (t²-0) + [tex]\frac{1}{6}[/tex] (t³ -0)
x = x₀ + vot - ½ c t² - 1/6 bt³
2) The speed when you reach the traffic light
Let's write the data that indicates, the initial velocity is vo = 15 m / s, the initial position is xo = 315m, let's use the initial values to find the constants.
t = 25 s x = 20
we substitute
20 = 315 + 15 25 - ½ c 25² - 1/6 b 25³
0 = 295 + 375 - 312.5 c - 2604.16 b
670 = 312.5 c + 2604.16 b
we simplify
2.144 = c + 8.33 b
Now let's use the equation for velocity,
v = v₀ - ct - ½ b t²
v = 15 - c 25 - ½ b 25²
v = 15 - 25 c - 312.5 b
let's write our two equations
2.144 = c + 8.33 b
v = 15 - 25 c - 312.5 b
Let's examine our equations, we have two equations and three unknowns (b, c, v) for which the system cannot be solved without another equation, in the statement it is not clear, but the most common condition is that if the semaphore does not change, it follows with this acceleration (constant) to a stop
a = c + b 25
from the first equation
c = 8.33 / 2.144 b
C = 3.885 b
we substitute in the other two
v = 15 - 25 (3.885 b) - 312.5 b
v = 15 - 409.6 b
final acelearation
a = 28.885 b
let's use the cinematic equation
[tex]v_{f}^2[/tex]= v² - 2 a x
0 = v² - 2a 20
0 = v² - (28.885b) 40
v² = 1155.4 b
we write the system of equations
v = 15 - 409.6 b
v² = 1155.4 b
resolve
v²= 1155.4 ( [tex]\frac{15 -v }{409.6}[/tex] )
v² = 2.8 ( 15 -v)
v² + 2.8 v - 42.3 = 0
v= [ -2.8 ±[tex]\sqrt {2.8^2 + 4 \ 42.3) }[/tex] ]/2 = [-2.8 ± 13.3]/2
v₁ = 5.25 m/s
v₂ = -8 m/s
A bump should primarily be used in which situation?
A. when the ball is going out of bounds
B. when returning a serve
C. to start game play
Force exerted on a body changes it's
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength ?
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;
[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]
Therefore, the electric field strength at the mid-point between the two rings is zero.
A ball on a string in uniform circular motion has a velocity of 8 meters per second, a mass of 2 kilograms, and the radius of the circle is 0.5 meters. What is the centripetal force keeping the ball in the circle?
Answer:
256 N
Explanation:
formula of centripetal force = mv²/r
m= 2kg
v= 8m/s
r= 0.5m
mv²/r = 2×8²/0.5 = 256N
If it takes 560 joules of work to move an object 10 meters what force was needed
Answer:
F = 56 N
Explanation:
Assuming that the applied force is parallel to the displacement, according to the definition of work, we can write the following expression for the force required:[tex]F = \frac{W}{\Delta x} =\frac{560J}{10m} = 56 N (1)[/tex]
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A train of waves is moving at a speed of 30 m/s. The frequency of the waves is 10 Hz. What is the wavelength?
a. 300 m
b. 30 m
c. 3 m
d. 0.1 m
Answer:
the answer is c, 3m......
When a train is moving at the speed of 30 m/s. The wavelength of the wave is 3 m.
What is wavelength?Wavelength is the characteristic of the wave which represents the distance between the adjacent crest or trough of the wave.
Given is the speed of the train of waves 30 m/s and frequency of the wave is 10 Hz, then the wavelength is related to speed and wavelength as
v = fλ
30 = 10 × λ
λ = 3 m
Thus, the wavelength of the wave is 3 m.
Learn more about wavelength.
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Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.6-V battery, the current from the battery is 2.07 A. When the resistors are connected in parallel to the battery, the total current from the battery is 8.98 A. Determine R1 and R2. (Enter your answers from smallest to largest.)
Answer:
When R1 = 2.193, R2 = 3.894
When R1 = 3.894, R2 = 2.193
Explanation:
We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.
Formula for resistance is;
R = V/I
R = 12.6/2.07
R = 6.087 ohms
Since R1 and R2 are connected in series.
Thus; R1 + R2 = 6.087 ohms
R1 = 6.087 - R2
We are also told that when they are connected in parallel, the current is 8.98 A.
Thus, R = 12/8.98
R = 1.403 ohms
Thus;
(1/R1) + (1/R2) = 1/1.403
Let's put 6.087 - R2 for R1;
(1/(6.087 - R2)) + (1/R2) = 1/1.403
Multiply through by 1.403R2(6.087 - R2) to get;
1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)
Expanding gives;
1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²
(R2)² - 6.087R2 + 8.54 = 0
Using quadratic formula, we have;
R2 = 2.193 ohms or 3.894 ohms
Thus,
R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894
R1 = 3.894 or 2.193
When R1 = 2.193, R2 = 3.894
When R1 = 3.894, R2 = 2.193
If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1x104 J of energy each second, how long does it take for the bridge’s oscillations to go from 0.1 m to 0.5 m amplitude?
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If soldiers march across the bridge with a cadence equal to the bridge's natural frequency and impart $$1.00 × 10^4 J$$ of energy each second, how long does it take for the bridge's oscillations to go from 0.100 m to 0.500 m amplitude. $ 5 \times 10^7 \text{J} $ . \\ b) $ 12 \times 10^4 \text{s}$ .
Your Welcome!
Explanation:
Which of the following have frequencies greater than orange light Your answer:
radio waves
purple light
ultraviolet rays
red light
green light
gamma rays
microwaves
infrared rays
Answer:
Gamma Rays have the highest frequencies
Explanation:
This is because Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies compared to the other light frequency. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation which means the answer has to be gamma rays. Brainly Please!!!! Here are screenshots that may help
(will give brainliest to whoever answers first and explains reasoning) A 10kg object is spun around in a circle with a centripetal acceleration of 3.5m/s^2. What is the centripetal force acting on the object?
Answer:
35 N
Explanation:
F = ma
centripetal force = 10(3.5) = 35 N
I don’t understand this
Answer:
true
Explanation:
force or powerbecause he pushes a disk
Solve the below problems being sure to provide the correct significant figures.
1) 1000 ÷ 4.886 = __________
2) 240 ÷ 12.3 = __________
3) 80 x 4.6 = __________
4) 4.527 x 30 = __________
5) 86 x 63.855 x 8000 = __________
6) 700 x 91.186 = __________
7) 7.1 x 348 = __________
8) 50 ÷ 29.1 = __________
9) 98.773 x 24.891 x 409 = __________
10) 0.065 x 3 x 3007 = __________
Answer:
1) 204.6663938
2) 19.51219512
3) 368
4) 135.81
5) 43932240
6) 63830.2
7) 2470.8
8) 1.718213058
9) 1005550.526
10) 586.365
Most of the questions you asked were in repeating decimal form.
Explanation:
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
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what colors of light are absorbed when white light falls on a green object?
1
Select the correct answer.
Which type of energy is thermal energy a form of?
A
chemical energy
B.
kinetic energy
C. magnetic energy
D. potential energy
Reset
Next
Answer:
B. kinetic energy
Explanation:
Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.
a boy throws a ball straight up into the air it reaches the highest point of its flight after 4 seconds how fast was the ball going when it left the boy's hand
Answer:
Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.
When6-2 He He-6 undergoes beta decay, the daughter is?
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah