Since, no external force is acting , so the system is in equilibrium .
Initial total energy = Final total energy
[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )
Therefore, option 1) is correct.
Hence, this is the required solution.
A man runs 400 meters north with a speed of 3 meters/second. How long does it take him to complete his trip? Answer in seconds. Round your answer to 3 decimal places. *
Answer:
133.333 s
Explanation:
From the question given above, the following data were obtained:
Displacement = 400 m North
Velocity (v) = 3 m/s
Time (t) =?
Thus, we can obtain the time taken for the man to complete his trip as shown below:
Velocity = Displacement /Time
3 = 400/time
Cross multiply
3 × time = 400
Divide both side by 3
Time = 400/3
Time = 133.333 s
Therefore, it will take the man 133.333 s to complete his trip.
Need help ASAP..please help
Answer:
option 3
Explanation:
can i get brainliest
A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. How far did the ball travel horizontally when it hit the ground?
414.9 m
500.76 m
510.2 m
130.5 m
Answer:
414.9 m
Explanation:
First, become familiar with the horizontal, and vertical vector components.
Vertical vector: Vy = V × sin (θ).
Horizontal vector: Vx = V × cos(θ).
Distance traveled = Velocity vector × time in the air.
Time in the air given Vy = 2 × Vy / g (in respect to the metric of the vector).
Range of the projectile = Vx² / g
Time in the air given Vx = (Vx + √(Vx)² + 2gh) / g.
Given a 28° angle with an initial velocity of 70m/s, we have enough information to calculate!
Vx = 70 m/s × cos(28°) ≈ 61.806 m/s
Vy = 70 m/s × sin(28°) ≈ 32.863 m/s
t = 2 × Vy / g
t = 2 × ≈32.863 / 9.8
t = ≈65.726 / 9.8
t ≈ 6.7 s
Distance traveled (horizontal) = Vx × t = 61.806 × 6.7 ≈ 414.9 m
At an instant when a particle of mass 80 g has a velocity of 25 m/s in the positive y direction, a 75-g particle has a velocity of 20 m/s in the positive x direction. What is the speed of the center of mass of this two-particle system at this instant?
a. 16 m/s
b. 45 m/s
c. 23 m/s
d. 20 m/s
e. 36 m/s
Answer:
16 m/s
Explanation:
Given that
y momentum = 0.080 *25 = 2
x momentum = 0.075*20 = 1.5
total momentum = √(4 + 2.25)
Total momentum = √6.25
Total momentum = 2.5
total mass = mass of x and y momentum = 0.075 + 0.080 = 0.155
speed of mass center = total momentum / total mass = 2.5/0.155 = 16.
And thus, the speed of the center of mass of this two-particle system at this instant is 16 m/s
5.List the four goals of Psychology. Give your own example for each one using a behavior
Answer:
describe, explain, predict, and change/control behavior.
Explanation:
describe: What are they doing? -Pavlov noticed that dogs were salivating when they would see his lab assistant before food was presented to them. This observation acted as a description of what was happening to them.
explain: Why are they doing that?- Pavlov started to look into why they were doing it. There was a stimulus, the assistant giving them food in the past to where they started to salivate at the sight of the lab assistant
predict: What would happen if I responded in this way?- Pavlov predicted that he could get the same reaction if he used a bell as a stimulus. Using this he was able to condition dogs to salivate at the ring of the bell.
change/control: What can I do to get them to stop doing that? Because of this discovery we can use conditioning today. For example, in the classroom teachers can use conditioning with their students to make it easier, parents to teach their children right from wrong and to have good behavior. (you do this bad thing you get time out, do a good thing and I will praise you, etc) It can be used when training employees and many other places.
A solid spherical ball and a hollow spherical ball made out of the same material are released from rest at the top of a ramp. They roll down the ramp without slipping to the bottom. On what quantities does the speed of each ball at the bottom of the ramp depend?A. Radius of the ball.B. Distribution of mass within the ball.C. Mass of the ball.D. Height of the ramp.
Answer:
D. Height of the ramp.
Explanation:
The solid spherical ball is expected to have more mass than that of the hollow spherical ball. And the speed of both balls would be influenced by the gravitational force as they roll down the ramp. Thus, the masses would move at different speed.
At the bottom of the ramp, the speed of the balls can be varied by varying the height of the ramp. So that the speed of both balls depend on the height of the ramp. As the height of the ramp increases, consequently, the speed of the balls increases. And if the height of the ramp decreases, the speed of the balls decreases consequently.
Find the net work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 54.0 N.
Answer:
The net work done by friction on the body of the snake is 333.35J
Explanation:
Work done is given by
W = F × s
Where W is the Work done
F is the force
and s is the distance covered
Since we are to determine the work done by friction, then we will determine the frictional force. The frictional force is given by
f = μN = μw
Where μ is the coefficient of friction
N is the normal reaction
and w is the weight
But, F = f
∴ W = μws
From the question
μ = 0.25
w = 54.0 N
Now, we will determine s
From the question,
We are to determine the work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius.
The distance s here is given by the circumference of the circle. Circumference of a circle is given by 2πr
∴ s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Hence,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
Hence, the net work done by friction on the body of the snake is 333.35J.
The net work done by friction on the body of the snake is :
-333.35J
FrictionFormulas:
Work done (W) = F × s
F = force
s = distance covered
f = μN = μw
μ = coefficient of friction
N = normal reaction
w = weight
Solution:
F = f
Weight is :
W = μws
μ = 0.25
w = 54.0 N
Distance covered:
s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Therefore,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
The net work done by friction on the body of the snake is 333.35J.
Learn more about "Friction":
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A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d, place a block of mass m on top of the compressed spring, and then release the block. The spring launches the block upward, and the block rises to a maximum height some distance above the now-relaxed spring. Find the speed of the block just as it loses contact with the spring.
Answer:
[tex]v=d\sqrt{\frac{k}{m}} [/tex]
Explanation:
In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:
[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]
In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:
[tex]U_{0}=K_{f}[/tex]
The initial potential energy of the spring is given by the equation:
[tex]U_{0}=\frac{1}{2}kd^{2}[/tex]
the Kinetic energy of the block is then given by the equation:
[tex]K_{f}=\frac{1}{2}mv_{f}^{2} [/tex]
so we can now set them both equal to each other, so we get:
[tex]=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}[/tex]
This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:
[tex]kd^{2}=mv_{f}^{2}[/tex]
so now we can solve this for the final velocity, so we get:
[tex]v=d\sqrt{\frac{k}{m}} [/tex]
Suppose a cart with no fans has a starting velocity of 2 m/s. What will be the velocity of the cart when it reaches the wall?
Answer:
less than stating velocity due to friction and air resistance.
Explanation:
A 3520 kg truck moving north makes an INELASTIC collision with an 1480 kg car moving 13.0 m/s east. After colliding, they have a velocity of 9.80 m/s at 66.9 degrees. What was the initial velocity of the truck? (m/s)
Answer:
v = 12.8 m/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components must be conserved too.Choosing a pair of axes coincident with the N-S and W-E directions, naming x to the W-E axis and y to the N-S one, we can write the following algebraic equations:[tex]p_{ox} = p_{fx} (1)[/tex]
[tex]p_{oy} = p_{fy} (2)[/tex]
Since we know all the information needed to solve (1), assuming a completely inelastic collision, we can focus in (2), writing both sides of the equation as follows:[tex]p_{oy} = m_{t} * v_{ot} = 3520 kg* v_{ot} (3)[/tex]
[tex]p_{fy} = m_{f} * v_{fy} = 5000 kg* 9.8 m/s * sin 66.9 = 45080 kg*m/s (4)[/tex]
Since (4) and (3) are equal each other, we can solve for vot, as follows:[tex]v_{ot} =\frac{45080kg*m/s}{3520kg} = 12.8 m/s (5)[/tex]
A 4.8-g particle is moving toward a stationary 7.4-g particle at 3.0 m/s. What percentage of the original kinetic energy is convertible to internal energy?
Answer:
60.185 percent of the original kinetic energy is convertible to internal energy.
Explanation:
Let suppose that collision between both particles is entirely inellastic. If there is no external forces exerted on any of the particles, then we can apply the Principle of Linear Momentum Conservation. That is:
[tex]m_{A}\cdot v_{A,o} + m_{B}\cdot v_{B,o} = (m_{A}+m_{B})\cdot v[/tex]
[tex]v = \frac{m_{A}\cdot v_{A,o}+v_{B}\cdot v_{B,o}}{m_{A}+m_{B}}[/tex] (1)
Where:
[tex]m_{A}[/tex] - Mass of the 4.8-g particle, measured in kilograms.
[tex]m_{B}[/tex] - Mass of the 7.4-g particle, measured in kilograms.
[tex]v_{A,o}[/tex] - Initial speed of the 4.8-g particle, measured in meters per second.
[tex]v_{B,o}[/tex] - Initial speed of the 7.4-g particle, measured in meters per second.
[tex]v[/tex] - Final speed of the collided particles, measured in meters per second.
If we know that [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex] and [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], then the final speed of the system is:
[tex]v = \frac{(4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)+(7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)}{4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg}[/tex]
[tex]v = 1.180\,\frac{m}{s}[/tex]
During the collision part of the initial energy is dissipated in the form of heat, which is related to the internal energy ([tex]\Delta U[/tex]), measured in joules. According to the Principle of Energy Conservation, we have the following model:
[tex]\Delta U = K_{A}+K_{B}-K[/tex] (2)
Where:
[tex]K_{A}[/tex], [tex]K_{B}[/tex] - Initial translational kinetic energies of each particle, measured in joules.
[tex]K[/tex] - Final translational kinetic energy of the collided particles, measured in joules.
By applying the definition of translational kinetic energy, we expand and simplify the equation above:
[tex]\Delta U = \frac{1}{2}\cdot m_{A}\cdot v_{A,o}^{2}+\frac{1}{2}\cdot m_{B}\cdot v_{B,o}^{2} -\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2}[/tex] (3)
If we get that [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]v = 1.180\,\frac{m}{s}[/tex], the internal energy associated with the system is:
[tex]\Delta U = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)^{2}+ \frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-\frac{1}{2}\cdot (4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg)\cdot \left(1.180\,\frac{m}{s} \right)^{2}[/tex]
[tex]\Delta U = 0.013\,J[/tex]
And the initial energy of both particles is:
[tex]E_{o} = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s}\right)^{2}+\frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}[/tex]
[tex]E_{o} = 0.0216\,J[/tex]
Lastly, the percentage of the original kinetic energy that is convertible to internal energy is: ([tex]\Delta U = 0.013\,J[/tex], [tex]E_{o} = 0.0216\,J[/tex])
[tex]\%e = \frac{\Delta U}{E_{o}}\times 100\,\%[/tex] (4)
[tex]\%e = \frac{0.013\,J}{0.0216\,J}\times 100\,\%[/tex]
[tex]\%e = 60.185\,\%[/tex]
60.185 percent of the original kinetic energy is convertible to internal energy.
A barge is 10 m wide and 60 m long and has vertical sides. The bottom of the boat is 1.2 m below the water surface. What is the weight of the barge and its cargo, if it is floating in fresh water of density 1000 kg/m^3?
a. 6.8 MN
b. 3.5 MN
c. 6.4 MN
d. 9.5 MN
e. 7.1 MN
Answer:
e. 7.1 MN approx.Explanation:
Step one:
given data
density of water= 1000kg/m^3
the dimension of the barge
width= 10m
length= 60m
depth of the boat in the water= 1.2m
Hence the volume occupied by the boat is
volume=10*60*1.2
volume= 720m^2
Step two:
Required is the weight of the barge
we can first find the mass using the relation
density = mass/volume
mass= density*volume
mass= 1000*720
mass= 720000kg
Step three:
Weight =mg
g=9.81m/s^2
W=720000*9.81
W=7063200N
divided by 10^6
W=7.06MN
W=7.1MN approx.
Which city did King David make the royal capital of the Kingdom of Israel?
Canaan
Jerusalem
Philistine
Cairo
Answer:
Jerusalem
Explanation:
because now a days jerusalem is the royal capital of the kingdom of israel so i conclude that the answer is jerusalem
At which stage of the scientific process should a scientific look at experimental data from other scientists to set up a new experimental study
Answer:
Hypothesis
Explanation:
A diffraction pattern is created on a screen when blue light is passed through a single slit. Does the central bright maximum in this pattern become wider or narrower, when the blue light is replaced by red light?
Answer:
becomes wider
Explanation:
From; asin θ= m λ
It was said in the question that in this particular instance blue light is replaced by red light. The wavelength of blue light is less than that of red light, it then follows that the central bright maximum in this pattern widens when red light is used.
Therefore the diffraction pattern becomes wider.
If the velocity of a car changes from 0 meters per second (m/s) to 100 m/s in 10 seconds, what is the acceleration over that 10 second period?
Answer:
10m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 100m/s
Time taken = 10s
Unknown:
Acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time.
A = [tex]\frac{v - u}{t}[/tex]
v = final velocity
u = initial velocity
t = time taken
So, insert the parameters and solve;
A = [tex]\frac{100 - 0}{10}[/tex] = 10m/s²
On a double-well roller coaster, the person starts from a height of 10.5 meters, the height of the middle bump of the well is 6.7 meters. How much kinetic energy does this person have at the middle of the middle bump from the bottom if the mass of the person is 79 kg? (g=9.8 m/s2) Round to two decimal places.
Answer:
"2941.96 J" is the appropriate answer.
Explanation:
The given values are:
Mass,
m = 79 kg
Height,
h = 10.5 m
Now,
The total energy at the given height will be,
⇒ Potential energy = mgh
On substituting the values, we get
= [tex]79\times 9.8\times 10.5[/tex]
[tex]=8129.1 \ J[/tex]
So,
⇒ At height 6.7 m, total energy = At height 10.5 m, total energy
On substituting the values, we get
⇒ [tex]Potential \ energy+Kinetic \ energy=8129.1[/tex]
⇒ [tex]79\times 9.8\times 6.7+Kinetic \ energy=8129.1[/tex]
⇒ [tex]5187.14+Kinetic \ energy=8129.1[/tex]
⇒ [tex]K.E=8129.1-5187.14[/tex]
⇒ [tex]K.E=2941.96 \ J[/tex]
Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.26 s. The period on the other planet turns out to be 3.82 s. What is the free-fall acceleration on that other planet?
Answer:
1.066 m/s^2
Explanation:
On the earth and on Mars respectively, the time periods of the pendulums are expressed as:
Tearth = 2pi * sqrt(L/gearth)
Tmars = 2pi * sqrt(L/gMars)
Divide the two equations above:
Tearth/Tmars = sqrt(gMars/gearth)
gMars = gearth(Tearth/Tmars)^2 = (9.80m/s^2)(1.26s/3.82s)^2 = 1.066 m/s^2
How far out from the sun is the rock line?
Answer:
5 million miles
Explanation:
an object of mass 4kg moving with the initial velocity of 20m/s accelerates for 10s and attaind a final velocity of 60m/s calculate the initial momentum
Explanation:
m = 4kg
u = 20m/s
t = 10s
v = 60m/s
initial momentum = mass × initial velocity
= 4 × 20 = 80kgm/s
A 4.5 kg puffer fish expands to 40% of its mass by taking in water. When the puffer fish is threatened, it releases the water toward the threat to move quickly forward. What is the ratio of the speed of the puffer fish forward to the speed of the expelled water backwards?
Answer:
2:5
Explanation:
We have the initial mass of the fish = 4.5kg
The mass of water intake = 40%of 4.5kg
= 0.40x4.5kg
= 1.8kg
The formula for conserving momentum
M1V1 = M2V2
M1 = mass of the puffer fish = 4.5kg
M2 = mass of the applied water = 1.8kg
V1 = velocity of the puffer fish
V2 = velocity of water
We are then to find v1/v2
V1/V2 = M2/M1
V1/V2 = 1.8/4.5
= 1/2.5
This is also 2/5
Therefore we have the ratio to be 2:5
Jared walks 120 m east, 150 m south, and then 40 m west. Find the total
distance traveled by Jared
Answer:
310 m
Explanation:
120+150+40=310
A falling stone takes delta t = 0.32s to travel past a window 2.2m Tall. From what height above the top of the window did the stone fall?
Answer:
The height above the top of the window is 1.44 m
Explanation:
Given;
time of motion, t = 0.32 s
height traveled at the given time, h = 2.2m
determine the initial velocity of the stone;
h = ut + ¹/₂gt²
2.2 = u(0.32) + ¹/₂ x 9.8 x 0.32²
2.2 = 0.32u + 0.502
0.32u = 2.2 - 0.502
0.32u = 1.698
u = 1.698 / 0.32
u = 5.31 m/s
This initial velocity on top of the window becomes the final velocity from the height above the window.
v² = u² + 2gh
where;
u is the initial velocity of the stone from the height above the window;
5.31² = 0 + (2 x 9.8)h
19.6h = 28.196
h = 28.196/19.6
h = 1.44 m
Therefore, the height above the top of the window is 1.44 m
24- What happens to a given mass of water as it is cooled from 4°C to zero?
(a) expands
(b) contracts
(c) vaporizes
(d) Neither expands, contracts, nor vaporizes.
Explanation:
A-It starts to expand. Hence ice and water of same weight has different volumes
Expansion happens to a given mass of water as it is cooled from 4°C to zero. Option A is correct.
What is temperature?Temperature directs to the hotness or coldness of a body. In straightforward terms,
it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles more the temperature.
Water loses density when it is cooled from 4°C to 0°C. It turns out that the temperature at which liquid water has the most significant density is 4 degrees Celsius.
It will increase in size when heated or cooled. Since most liquids shrink when they are chilled, it is rare for water to expand when cooled to lower temperatures.
As a mass of water is cooled from 4°C to zero, expansion occurs.
Hence, option A is correct.
To learn more about the temperature refer to the link;
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An ideal gas expands quasi-statically and isothermally from a state with pressurepand volumeVto a state with volume 4V. How much heat is added to the expanding gas?
Answer:
Q = PV(In 4)
Explanation:
We are told that the volume expands from V to a state with volume 4V.
Thus, initial volume is V and Final volume is 4V.
We want to find How much heat is added to the expanding gas.
For an isothermal process, the work done is calculated from;
W = nRT(In(V_f/V_i))
Where;
V_f is final volume
V_i is initial volume
Thus;
W = nRT(In(4V/V))
W = nRT(In 4)
Now, from ideal gas equation, we know that;
PV = nRT
Thus;
W = PV(In 4)
Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W
Where Q is quantity of heat
Thus;
Q = PV(In 4)
A string is waved up and down to create a wave pattern with a wavelength of 0.5 m. If the waves are generated with a frequency of 2 Hz, what is the speed of the wave that travels through the string to the other end? The speed of the wave is __ m/s.
Answer:
1m/s
Explanation:
Speed of a wave is expressed as;
Speed = frequency × wavelength
Given
Frequency = 2Hz
Wavelength = 0.5m
Required
Speed of the wave
Substitute
Speed = 2×0.5m
Speed = 1.0m/s
Hence the speed is 1.0m/s
A repeated back and forth or up and down motion is called a
Answer:
A vibration is a repeated back-and-forth or up-and-down motion.
Explanation:
Waves carry energy through empty space or through a medium without transporting matter.
A typical sheet of printer paper is 8.5 inches wide and 11 inches long.
a.) Calculate the length of one sheet in cm
b.) Calculate the width of one sheet in cm?
Answer:
Explanation:
Given the length and width in inches
Length = 8.5in
Width = 11in
b) Convert to centimeters
1in = 2.54cm
8.5in = x
Cross multiply
x = 8.5 × 2.54
x = 21.59cm
Hence the width in cm is 21.59cm
a) 11in = y
1in = 2.54cm
Cross multiply
x = 11 × 2.54
x = 27.94cm
Hence the length in cm is 27.94cm
Which computational model can be applied to the total internal energy of a system with two point masses?
{Energy Flows Quick Check}
Answer:
D
Explanation:
Took the test :)
A rock is thrown straight upward with a speed 6.000 m/s. If this rock takes 1.540 s to fall to the ground, from what height h was it released?
Answer:
3.8961038961038
Explanation:
6000/1540=3.8961038961038
X= v/t