Answer:
distance = 112 miles
Explanation:
its 12 miles every 0.6 in a hour
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?
A red pool ball is rolling directly east before it collides with the
white cue ball moving directly north. Due to conservation of
momentum the total momentum of both objects after the
collision would be in which direction?
Answer: North East
Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.
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
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 train travels at a speed of 70km/h and travels a distance of 630 km. How long did it take the train to complete its journey?
Answer:
9 hr
Explanation:
Speed = distance/time
Let the time taken by the train be t.
=> 70 = 630/t
=> t = 630/70
=> t = 9
it take 9 hr to complete its journey
Answer:
9 hours
Explanation:
Using the formula to solve motion problems: d=r*t (Distance = Rate * Time)
Given:
Distance = 630 km
Rate (Speed) = 70 km/h
Time = Distance / Time
Time = 630/70
= 9
An 5 kg object moving at 10 m/s will have a momentum equaling ____________.
15 kg m/s/s
15 kg m/s
Answer:
50Kgm/s
Explanation:
Momentum=Mass*Velocity
P=mv
Given Mass=5Kg. Given Velocity=10m/s
Momentum=5*10=50Kgm/s
A small dog is trained to jump straight up a distance of 1.1 m. How much kinetic energy does the 7.7 kg dog need to jump this high?
Answer:
83.09 J
Explanation:
The potential energy at the point of the top of the jump is represented by the equation
[tex]mgdeltah[/tex]
when the dog jumps, all the potential energy converts to kinetic energy (1/2mv^2). Plugging in the values:
(7.7)(4.184)(1.1) = 83.0907 J
Is there a way to see moon and the sun at once?
The rods, which number over 100 million, can only be activated by a certain range of wavelengths, but they do not pass any color information to the brain. In other words, they note differences in shades of grey (from black to white) and are responsible for a person's ability to see in dim light. Cones, which number around 6 million, give us color vision. Cones come in three different kinds: 64%% of cones are sensitive to long wavelengths of visible light (toward the red end of the spectrum), 32%% are sensitive to medium wavelengths, and the remaining 2%% are sensitive to short wavelengths (toward the blue end of the spectrum). Colors are differentiated on the basis of the extent to which visible light stimulates each kind of cone.
Do rods have their peak sensitivity at a higher or lower frequency than cones?
a) Higher
b) Lower
Answer:
The correct answer is a
peak sensitivity is much higher for cones
Explanation:
After reading this interesting problem, where it gives a good description of the types of photoreceptor cells that exist in the eyes
The Cone has its name because of the shape of a cone that has this shape that allows to perceive very small amounts of intensity
The Canes have the shape of a cane and are filled with a substance that is sensitive to color colors, but they need a greater intensity of light to be activated, for which reason they work in the daytime, when it gets dark the intensity of the light is insufficient to activate these cells and the only ones that send information to the brains are the cones.
With this explanation it is clear that cones have high sensitivity at all frequencies and rods have low to medium sensitivity at specific frequencies.
Therefore peak sensitivity is much higher for cones
The correct answer is a
You exert a 138 N push the leftmost of two identical blocks of mass 244 g connected by a spring of stiffness 605 kg/s2. After pushing the block a distance 15 cm, you release it; by this time the rightmost block has moved a distance 5 cm. (a) What is the energy in the oscillations between the blocks
Answer:
the energy in the oscillations between the blocks is 3.025 J
Explanation:
Given the data in the question;
Force f = 138 N
stiffness of spring k = 605 kg/s²
mass of block = 202 g = 0.202 kg
pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm
i.e
x₁ = 15 cm
x₂ = 5cm
the energy in the oscillations between the blocks will be;
E[tex]_A[/tex] = E[tex]_B[/tex] = [tex]\frac{1}{2}[/tex]k( Δx )²
we substitute
= [tex]\frac{1}{2}[/tex] × k( 15 - 5 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × ( 10 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × 100 × 10⁻⁴
= 3.025 J
Therefore, the energy in the oscillations between the blocks is 3.025 J
I don’t understand this
Answer:
true
Explanation:
force or powerbecause he pushes a disk
What is the magnitude of velocity for a 3,100 kg car possessing 4,100 kg•m/s of momentum?
Answer:
my c-ock and balls be rotating im gunna be hxrny all day every day im gunna f-u-ck u and r-ape ur family girls should go back to the kitchen and hope not to be beat by their husband uh yea mhm girls desurve to be r-aped f-u-ck buck suck c-u-ck
Explanation:
Answer:
3.75
Explanati
At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B
Answer:
0.375
Explanation:
For incompressible flow, we know that;
ρ1•v1•A1 = ρ2•v2•A2
Where;
ρ1 = density of fluid at position A
v1 = speed of fluid at position A
A1 = area of tube
ρ2 = density of fluid at position B
v2 = speed of fluid at position B
A2 = area of tube
We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.
Thus;
ρ1/ρ2 = (v2•A2)/(v1•A1)
Now, the tube will have the same height.
But we are given;
diameter of A = 12.00 cm = 0.12 m
diameter of B = 6 cm = 0.06 m
Thus;
A1 = π(d²/4)h = πh(0.12²/4)
A2 = πh(0.06²/4)
We are also given;
v1 = 12 m/s
v2 = 18 m/s
Thus;
ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))
πh/4 will cancel out to give;
ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)
ρ1/ρ2 = 0.375
23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
1. A horizontal force of 50 N is applied to push a desk 40 m across a
warehouse floor. How much work is done?
2000 J
1000 J
3000 J
2001 J
Two students Tim and Alane travel to South Dakota. Tim stands on Earth’s surface and enjoys some sunshine. At the same time, Alane descends into a gold mine where neutrinos are detected, Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences.
Answer:
Explanation:
Neutrinos are otherwise called leptons. They are principal particles. A lepton is a rudimentary half-spin molecule that doesn't go through solid reactions. Neutrinos are not usually charged and exceptionally light weighted so they once in a while interface with other matter. Neutrinos are light weighted. Their mass is around 10⁻⁷ kg. A neutrino possesses a small radius, too little to ever be estimated. A little span and very less mass make them imperceptible. Since neutrinos have next to no mass. they travel at almost the speed of light and thus they arrive at the outside of the Sun in only 2 seconds, dissimilar to photons which take convoluted ways to arrive at the Sun's surface in a huge number of years.
The photon and neutrino, both were made in the Sun's center yet on various occasions. The neutrino is only a couple of minutes old though the photon is around 1,000,000 years of age. At the point when the photon was made in the Sun's center. it needed to venture out to the outside of the Sun. in any case, rather because of its hefty mass and cooperation with other matter, it headed out a crisscross way to the surface. Ordinarily, it was repulsed and it was sent back to the middle where it needed to begin once more. It required a large number of years for a photon to arrive at the outside of the Sun.
Nonetheless, when it arrived at the Sun's surface, it required just 8.8 minutes for the photon to arrive at Earth. The neutrino was anyway made only a couple of minutes prior in the Sun's center. Since it has an entirely irrelevant mass, little size, and no charge, it didn't interface with its environmental factors. So it just required 2 seconds for the neutrino to arrive at the Sun's surface. When it arrived at the Sun's surface, it arrived at the earth in about 8.8 minutes. with the photon. So both, photon and neutrino have various histories as the two of them were made at a hole of around 1,000,000 years.
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.
an inventor makes a clock using a brass rod and a heavy mass as a pendulum.WHAT Happens when the clock get colder?
An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time
What is thermal expansion?The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.
As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.
The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum
T = 2π√(L/g)
As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time
Thus, if a pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder
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Diffraction occurs for all types of waves, including sound waves.
a. True
b. False
Answer:
a. True
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.
Mathematically, the speed of a sound is given by the formula:
[tex] Speed = wavelength * frequency [/tex]
Generally, the frequency of a sound wave determines the pitch of the sound that would be heard.
Diffraction occurs for all types of waves, including sound waves.
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Please why does a simple pendulum graph fail to pass through the origin ?
Answer:
The friction of the string and its pivotal anchor point cannot be eliminated. The precise measurement of the length of the pendulum is difficult to take by using meter sticks or rulers. The value of the acceleration due to gravity g in the locality is not constant and must be obtained from reliable sources.
You cannot completely remove the friction between the string and its key anchor point. Using meter sticks or rulers makes it challenging to take an accurate measurement of the pendulum's length. Since it is not constant, it is necessary to get accurate information on the local gravity acceleration related to g.
What is a pendulum?The body is suspended from a central object to act as a pendulum that swings back and forth as a result of gravity. Because the period—the amount of time between each full oscillation—is constant, pendulums are employed to control the movement of clocks. The formula for a pendulum's period T is T = 2 Square root of L/g, where L is the pendulum's length and g is the acceleration brought on by gravity.
Galileo, an Italian scientist, made the first observation about the consistency of a pendulum's period by contrasting the motion of a swinging lantern in a cathedral in Pisa with his heart rate.
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In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.5 m from his foot at the edge of the pool.
Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2.1 m deep pool?
Answer:
4.4 m
Explanation:
We are told the light from his flashlight, 1.3 m above the water level. Thus; h1 = 1.3m
Also,we are told that the light shone 2.5 m from his foot at the edge of the pool. Thus, L1 = 2.5 m
Angle of incidence θ1 is given by;
tan θ1 = L1/h1
tan θ1 = 2.5/1.3
tan θ1 = 1.9231
θ1 = tan^(-1) 1.9231
θ1 = 62.53°
Using Snell's law, we can find the angle of refraction from;
Sin θ2 = (η_air/η_water) Sin θ1
Where;
η_air is Refractive index of air = 1
η_water is Refractive index of water = 1.33
Thus;
Sin θ2 = (1/1.33) × sin 62.53°
Sin θ2 = 0.6671
θ2 = sin^(-1) 0.6671
θ2 = 41.84°
We want to find where the spot of light hit the bottom of the pool if the pool is 2.1 m deep. Thus, h2 = 2.1 m
Now, the spot can be found from;
L = L1 + L2
Where L2 = (h2) tan θ2
L = 2.5 + 2.1 tan 48.84
L = 2.5 + (2.1 × 0.8954)
L ≈ 4.4 m
ASAP Even though the force exerted on each object in a collision is the same strength, if the objects have different masses, their will be different. * O changes in velocity O amount of force O speed and direction
Answer:
it should be changes in velocity
Explanation:
I hope this helps!
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?
Answer: Hello, Mark me as Brainliest! :)
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:
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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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SIMPLE HARMONIC MOTION 1.0 Objective To study simple harmonic motion by observing the motion of a simple pendulum. 2.0 Simple harmonic motion Oscillatory motion is extremely common in nature. Examples include waves (water, sound, earthquake, etc.) and vibrations produced by musical instruments. If the oscillation is characterized by a constant frequency and amplitude (if the motion reproduces itself in a fixed time period T), then the motion is said to be "harmonic." If the oscillation can be described as a sinusoidal function of time and position, the motion is said to be "simple harmonic." Simple harmonic motion (SHM) occurs when for every applied force or torque, there is a restoring force or torque which is proportional to the displacement of the system from its equilibrium position. 2.0.1 Name two real-world examples of harmonic motion. (Hint: Southern California is known for what sort of natural disasters?)
Answer:
the waves in the sea, leaves of the trees, cables in the bridges, pendulum clock
Explanation:
In nature there are many examples of simple harmonic motion, for example.
* The movement of the waves in the sea is an oscillation movement up and down
* The movement of the leaves of the trees when a wind blows and then stops, but the leaf and branches are oscillating
* The movement of the cables in the bridges, especially in the suspension bridges
* The movement of a pendulum clock
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.
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.
Answer:
Δθ = 15747.37 rad.
Explanation:
The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
Solving for Δθ in (2):[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N