Answer: 117.6N
Explanation:
By the second Newton's law, we know that:
F = m*a
F = force
m = mass
a = acceleration
We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.
Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:
F = 12kg*9.8m/s^2 = 117.6N
How much work would be done on a particle with 5.0 C of charge on it if it moved from an equipotential line at 5.5 volts to another equipotential line at 3.5 volts?
Answer:
10J
Explanation:
In this question we have the following information
The charge of the particle is q = 5 C
The equipotenetial level is V1 = 5.5 v
and also the
equipotenetial level is V2 = 3.5 v
So we calculate the
work done W=q x (v1-v2)
workdone = 5 x (5.5-3.5)
= 5x2
=10 J
Workdone = 10 J
So we conclude that the workdone on a particle with these information is 10j
How much work is done by the gravitational force on the block?
Answer:
Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.
Explanation:
True or False when an object speeds up it gains momentum
Answer: True
Explanation:
A school bus drives north and then east through the city as it takes students to school. The bus crosses a city block every 10 seconds. If all the city blocks are the same length, what can be said about the motion of the bus?
Answer:
The bus is moving at a constant speed.
Explanation:
We have the following facts from the question;
- The bus crosses a city block every 10 seconds.
- all the city blocks are the same length
Since all the city block are the same length, let's say the distance is d.
Since it crosses the city block every 10 seconds, we know that;
Speed = distance/time
Thus, Speed = d/10
Thus,it is moving at this speed of d/10 all through.
Therefore we can conclude that the bus is moving at a constant speed.
Assume a hockey player accelerates from 0 ft/s to 24 ft/s over a period of 2
seconds. What is his acceleration if his direction does not change? Explain in
words what this acceleration means.
Answer:
12ft/s or 4m/s
Explanation:
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?
Answer:
Explanation:
m1v1=m2v2
m1=70 kg
m2=10 g=0.01 kg
v2=500 m/s
m1v1=m2v2
v1=m2v2/m1
v1=0.01*500/70
v1=0.07
What would happen if there is more male hyenas than female hyenas in a population?
Choices:
Male hyenas will compete to mate with the females.
Some male hyenas will die.
Male hyenas for wait for more females to join the population.
Answer:
Option 1
Explanation:
I always see animals do that
Which temperature is warmer than the freezing point of water?
O A. OK
O B. 33K
O C. 1°C
O D.O°F
Answer:
C 1 degree
Explanation:
Is Natural Gas nonrenewable or renewable? Why? Use in your own words.
Answer:
Natural gas is non renewable energy.
Explanation:
Because they were formed from the buried reamains of plants and animals that live a million years ago. It is formed from fossil fuels.
A step-down transformer converts 120 V to 5 V in a phone charger. If the primary coil has 600 turns, how many turns are in the secondary coil? Show all work with correct answer and unit as well.
Answer:
25 turns
Explanation:
Given the following data;
Input voltage = 120V
Output voltage = 5V
Number of turns in primary coil = 600 turns
[tex] \frac {Ns}{Np} = \frac {Vs}{Vp} [/tex]
Where;
Ns is the number of turns on the secondary coil. Np is the number of turns on the primary coil. Vs is the voltage across the secondary coil (output voltage). Vp is the voltage across the primary coil (input voltage).Substituting into the equation, we have;
[tex] \frac {Ns}{600} = \frac {5}{120} [/tex]
Cross-multiplying, we have;
[tex] 120*Ns = 600*5[/tex]
[tex] 120Ns = 3000[/tex]
[tex] Ns = \frac {3000}{120}[/tex]
Ns = 25 turns
Therefore, the number of turns in the secondary coil is 25 turns.
Hence, it is a step-down transformer because the number of turns on the primary coil is greater than the number of turns on the secondary coil. Consequently, the output voltage (5V) will be lesser than the input voltage (120V).
An 7.40 kg block drops straight down from a height of 0.83 m, striking a platform spring having a force constant of 9.50 102 N/m. Find the maximum compression of the spring.
Answer:
0.25 m.
Explanation:
mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.
The maximum compression on the spring can be determined by;
Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
But, potential energy = mgh
So that,
mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]
= 0.06336
x = 0.2517
x = 0.25 m
The maximum compression of the spring is 0.25 m.
. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.
Answer:
The maximum speed of the car is 35 m/s
The total distance traveled by the car is 658.33 m
Explanation:
Given;
initial velocity of the car, u = 15 m/s
acceleration of the car, a = 2 m/s²
time of car motion, t = 10 s
(i)
Initial distance traveled by the car is given by;
d₁ = ut + ¹/₂at²
d₁ = (15 x 10) + ¹/₂(2)(10)²
d₁ = 150 + 100
d₁ = 250 m
The maximum speed of the car during this is given by;
v² = u² + 2ad₁
v² = (15)² + (2 x 2 x 250)
v² = 1225
v = √1225
v = 35 m/s
(ii)
The final distance cover by the car during the deceleration of 1.5 m/s².
Note: the final or maximum speed of the car becomes the initial velocity during deceleration.
v² = u² + 2ad₂
where;
v is the final speed of the car when it stops = 0
0 = u² + 2ad₂
0 = (35²) + (2 x - 1.5 x d₂)
0 = 1225 - 3d₂
3d₂ = 1225
d₂ = 1225 / 3
d₂ = 408.33 m
The total distance traveled by the car is given by;
d = d₁ + d₂
d = 250 m + 408.33 m
d = 658.33 m
You are hanging on to the edge of a merry-go-round, and must exert a force of 100 N to hang on. If the speed of the merry-go-round doubles, how much force will you need to exert to hang on?
Answer:
If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
Explanation:
Given;
initial force exerted to hang on, F₁ = 100 N
The force exerted on the merry-go-round in order to hang on must be an inward force known as centripetal force.
Centripetal force is given by;
[tex]F_c = \frac{mv^2}{r} \\\\keeping \ "m" \ and \ "r" \ constant, we \ will \ have \ the \ following \ equation;\\\\\frac{F_c_1}{v_1^2} = \frac{F_c_2}{v_2^2} \\\\F_c_2 = \frac{F_c_1*v_2^2}{v_1^2}\\\\when \ the \ speed\ doubles \ i.e, v_2 = 2v_1\\\\ F_c_2 = \frac{F_c_1*(2v_1)^2}{v_1^2}\\\\ F_c_2 = \frac{F_c_1*4v_1^2}{v_1^2}\\\\F_c_2 = F_c_1 *4\\\\F_c_2 = 4(F_c_1)\\\\F_c_2 = 4 (100 \ N)\\\\F_c_2 = 400 \ N[/tex]
Therefore, If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
A solid sphere rolling without slipping on a horizontal surface. If the translational speed of the sphere is 2.00 m/s, what is its total kinetic energy?
Answer:
The total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)
Explanation:
The total kinetic energy of a sphere is given by the sum of the rotational kinetic energy and the translational kinetic energy. That is,
[tex]K_{Total} = K_{R} + K_{T}[/tex]
The rotational kinetic energy [tex]K_{R}[/tex] is given by
[tex]K_{R} = \frac{1}{2}I\omega^{2}[/tex]
Where [tex]I[/tex] is the moment of inertia
and [tex]\omega[/tex] is the angular velocity
The translational kinetic energy [tex]K_{T}[/tex] is given by
[tex]K_{T} = \frac{1}{2}mv^{2}[/tex]
Where [tex]m[/tex] is the mass
and [tex]v[/tex] is the translational speed (velocity)
∴ [tex]K_{Total} = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}[/tex]
But, the moment of inertia [tex]I[/tex] of a sphere is given by
[tex]I = \frac{2}{5}mr^{2}[/tex]
Where [tex]m[/tex] is mass
and [tex]r[/tex] is radius
∴ [tex]K_{Total} = \frac{1}{2}\times \frac{2}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}[/tex]
[tex]K_{Total} = \frac{1}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}[/tex]
Also, [tex]\omega = \frac{v}{r}[/tex]
∴ [tex]\omega^{2} = \frac{v^{2} }{r^{2} }[/tex]
Then,
[tex]K_{Total} = \frac{1}{5}mr^{2} \times \frac{v^{2} }{r^{2} } + \frac{1}{2}mv^{2}[/tex]
[tex]K_{Total} = \frac{1}{5}mv^{2} + \frac{1}{2}mv^{2}[/tex]
∴ [tex]K_{Total} = \frac{7}{10}mv^{2}[/tex]
From the question, [tex]v = 2.00 m/s[/tex]
Then,
[tex]K_{Total} = \frac{7}{10}m(2.00)^{2}[/tex]
[tex]K_{Total} = \frac{7}{10}m\times 4.00[/tex]
[tex]K_{Total} = 2.8m J[/tex]
Hence, the total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)
Calculate the gravitational potential energies of the melon and the pomegranate. Which one has potential energy
Answer:
The mellon
Explanation:
I just know
A scuba diver wears weights as well as a buoyancy compensator to establish neutral buoyancy while diving. The buoyancy compensator can either be inflated with air or the air in it can be released. Explain how a scuba diver uses the buoyancy compensator to dive and to rise back to the surface.
Answer:
With more air is more buoyancy. When deflated or released the scuba diver is less buoyant.
Explanation:
The compensator is a Buoyancy control device that has an inflatable air bladder.When we have more air out into the inflatable bladder, then one is more buoyant. If the air is released from the bladder, then one is less buoyant. We add air through an air inflation valve. Air is also then released using air-deflation valves.
Buoyancy can be defined as an upward force which is exerted on an object that is fully or partially immersed in water
when one is less buoyant than water, it means that the upward pressure is more than the downward pressure of that person and his equipment. Then he will float. In a case of negative buoyancy, we have downward pressure of this person and his equipment to be more than the upward pressure of the water. Then sinking will happen.
A bug crawls 2.25 m along the base of a wall. Upon reaching a corner, the bugs direction of travel changes from south to west. THe bug that crawls 3.15 m before stopping. What is the magnitude of the bugs displacment?A) 5.40 m.B) 2.72m.C) 3.45 m.D) 3.87 in.E) 4.29 m.
Answer:
The magnitude of the bugs displacement is 3.87 m
Explanation:
An illustrative diagram for the scenario is given in the attachment below.
In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is
x² = 2.25² + 3.15²
x² = 5.0625 + 9.9225
x² = 14.985
x = √14.985
x = 3.87 m
Hence, the magnitude of the bugs displacement is 3.87 m.
A wave has a period 2.00 s, an amplitude 20.0 cm, and a wavelength 3.00 m. What is the speed of the wave?
a. 10.0 cm/s
b. 0.100 cm/s
c. 340 m/s
d. 0.667 m/s
e. 1.50 m/s
Answer:
a)
Explanation:
because if in 2 seconds have 20.0 cm/2s cm in 1 second have 10.0 cm/s
What does fitness means to you
Answer:
Exercising, yoga
Explanation:
10- The coefficient of volumetric expansion for gold is 4.20 x 10°/C°. The density of gold is
19 300 kg/m at 0.0 °C. What is the density of gold at 1050 °C?
(a) 20 200 kg/m
(b) 19 000 kg/m
(c) 18 500 kg/m
(d) 19 300 kg/m
(e) 18 800 kg/m
Answer:
ρ = 19215 kg / m³ , the answer the correct is D
Explanation:
All materials are sold when heated, in first approximation
ΔV = β V₀ (T -T₀)
for this case
ΔV = 4.20 10⁻⁶ Vo (1050 - 0)
ΔV = 4.41 10⁻³ Vo m³
V-Vo = 4.41 10⁻³ Vo
V = 1.00441 Vo
density is defined by
ρ = m / V
a T = 0ºc
ρ₀ = m / Vo
aT= 1050ºC
ρ = m / V
ρ = m / (1,00441 Vo)
ρ = 1 / 1.00441 m/Vo
ρ = 0.9956 ρ₀
ρ = 0.9956 19300
ρ = 19215 kg / m³
when checking the answers the correct one is D
What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical position to its highest vertical position?
Answer:
ΔU = 2 mg h
Explanation:
In a spring mass system the potential energy is U = m g h
where h is measured from the equilibrium point of the spring
the potential energy at the highest point is
U₁ = m g h
the potential energy at the lowest point is
U₂ = m g (-h)
instead in this energy it is
ΔU = 2 mg h
In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero
Potential energy is defined as the energy stored in a body which may convert into kinetic energy when moved.
The Formule of the potential energy is [tex]mgh[/tex] The correct answer is 0
Hence, [tex]U = 2 mg h[/tex]
The H stated as the measured from the equilibrium point of the spring
Therefore, the potential energy at the maximum point is [tex]U_1 = m g h[/tex] and the potential energy at the minimum point is [tex]U_2 = m g (-h)[/tex]
Hence, after solving it we got
[tex]U_1 -U_2 = 2 mg h[/tex]
Therefore, the energy change in the process is 0.
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7. It is the art of drawing solid objects on two-dimensional surfaces.
Explanation:
Perspective- the art of drawing solid objects on a two-dimensional surface so as to give the right impression of their height, width, depth, and position in relation to each other when viewed from a particular point.
If Earth’s Moon were replaced with a typical neutron star, what would the angular diameter of the neutron star be as seen from Earth?
Answer:
[tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex]
Explanation:
d = Diameter of typical neutron star = 20 km = 20000 m
D = Distance between Earth and Moon = [tex]384.4\times 10^6\ \text{m}[/tex]
Here, [tex]D>>d[/tex] so we use small angle approximation
[tex]\delta=\dfrac{d}{D}\\\Rightarrow \delta=\dfrac{20000}{384.4\times 10^6}\\\Rightarrow \delta=0.00005202\ \text{rad}=\dfrac{0.00005202\times 180}{\pi}=0.003^{\circ}[/tex]
The angular diameter of the neutron star would be [tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex] from Earth.
Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) What is the change in the temperature of the gas?________ ? K(b) Find the change in its internal energy.________ ? J(c) Determine the change in pressure.________ ? Pa
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
A) The change in the temperature of the gas is; ΔT = 139.5 K
B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J
C) The change in pressure of the gas is; ΔP = 2899.5 Pa
We are given;
Volume of Monatomic ideal gas; V = 1.2 m³
Amount of heat added; Q = 5.22 × 10³ J
number of moles; n = 3
A) To calculate the change in temperature of the monatomic idea gas, we will use formula;
Q = ³/₂nRΔT
Where R is a constant = 8.314 J/mol.K
ΔT is the change in temperature
Making ΔT the subject of the formula;
ΔT = ²/₃(Q/(nR))
ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)
ΔT = 139.5 K
B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;
ΔU = Q
Thus;
change in internal energy; ΔU = 5.22 × 10³ J
C) The change in pressure will be calculated from the formula;
ΔP = (n*R*ΔT)/V
ΔP = (3 * 8.314 * 139.5)/1.2
ΔP = 2899.5 Pa
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A sample of gas, initially with a volume of 1.0 L, undergoes a thermodynamic cycle. Find the work done by the gas on its environment during each stage of the cycle described below. (Enter your answers in J.)
Answer:
(a) W₁ = 3293.06 J = 3.293 KJ
(b) W₂ = 0 J
(c) W₃ = - 506.625 J = - 0.506 KJ
(d) W₄ = 0 J
(e) W = 2786.435 J = 2.786 KJ
Explanation:
The complete question has following parts:
(a) First gas expands from a volume of 1 L to 6 L at a constant pressure of 6.5 atm
(b) Second, the gas is cooled at constant volume until the pressure falls to 1 atm
(c) Third, the gas is compressed at a constant pressure of 1 atm from a volume of 6 L to 1 L.
(d) Finally the gas is heated until its pressure from 1 atm to 6.5 atm at constant volume
(e) what is the net work?
ANSWERS:
(a)
The work done by a gas at constant pressure is given as follows:
W = PΔV
where,
W = Work done by the gas
P = Constant Pressure of the Gas
ΔV = Change in Volume of The gas
Therefore, for the first step:
P = P₁ = (6.5 atm)(1.01325 x 10⁵ Pa/1 atm) = 6.58613 x 10⁵ Pa
ΔV = ΔV₁ = 6 L - 1 L = (5 L)(0.001 m³/1 L) = 5 x 10⁻³ m³
W = W₁
Therefore,
W₁ = (6.58613 x 10⁵ Pa)(5 x 10⁻³ m³)
W₁ = 3293.06 J = 3.293 KJ
(b)
The work done at a constant volume by a gas is always zero due to no change in volume:
W₂ = P₂ΔV₂
W₂ = P₂(0)
W₂ = 0 J
(c)
For the third step:
P = P₃ = (1 atm) = 1.01325 x 10⁵ Pa
ΔV = ΔV₃ = 1 L - 6 L = (- 5 L)(0.001 m³/1 L) = - 5 x 10⁻³ m³
W = W₃
Therefore,
W₃ = (1.01325 x 10⁵ Pa)(-5 x 10⁻³ m³)
W₃ = - 506.625 J = - 0.506 KJ
(d)
The work done at a constant volume by a gas is always zero due to no change in volume:
W₄ = P₄ΔV₄
W₄ = P₄(0)
W₄ = 0 J
(e)
Hence, the net work is given as follows:
W = W₁ + W₂ + W₃ + W₄
W = 3293.06 J + 0 J + (- 506.625 J) + 0 J
W = 2786.435 J = 2.786 KJ
Suppose you put an ice cube into a cup of hot tea. In what direction does energy in the form of heat flow? What happens to the ice cube as this flow of energy occurs?
Answer:
The energy flows between the ice and the tea equally. The table below shows the temperatures of several different objects made of the same material.
Which describes the genetic disorder that causes neurons in the brain to break down? scribes the genetic disorder that causes neurons in the brain to break down need help will mark brainliest
Genetic disorders and Parkinson's disease.
Define Hydrostatic Equilibrium
Answer:
well, in my view,
In fluid mechanics, hydrostatic equilibrium or hydrostatic balance (also known as hydrostasy) is the condition of a fluid or plastic solid at rest. This occurs when external forces such as gravity are balanced by a pressure-gradient force.
David Wetterman drops a 5 kg watermelon from the top of a 30 m building. What is the velocity of the watermelon as it smashes
into the ground (neglecting air resistance)?
-(1)
A)
24.25 m/s
B)
32.45 m/s
C)
60 m/s
D)
588 m/s
Answer:
A. 24.25 m/s
Explanation:
velocity = [tex]\sqrt{2 * g * d}[/tex]
velocity = sqr 2 * 9.8 * 30 = sqr 588 = 24.25 m/s
The velocity of the watermelon as it smashes into the ground will be 24.2 m/s
State the third equation of motion?
The third equation of motion is -
v² - u² = 2aS
Given David Wetterman drops a 5 kg watermelon from the top of a 30 m building.
Height of building [S] = 30 m
Mass of watermelon [M] = 5 Kg
Initial velocity [v] = 0 m/s
acceleration [g] = 9.8 m/s²
Using the third equation of motion -
v² - u² = 2aS
v² = 2aS
v² = 2 x 9.8 x 30
v² = 588
v = 24.2 m/s
Therefore, the velocity of the watermelon as it smashes into the ground will be 24.2 m/s.
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Give real-world examples of evidence that supports the evolution of Earth in each category:
Deposition -
Chemical weathering -
Volcanic eruption -
Answer:
concave shape of the waverock
Explanation:
The wave rock is formed by the weathering of the surrounding area. This helps in proving the deposition part, as the wave rock was below the ground, occurred due to deposition of rock years over the years. It is made from very tough material from its surroundings. The weathering reduces the surrounding terrain, while the bedrock remains to witness for the history. Also, volcanic eruptions have been changing earth for a long time. (first of all, the theory that a volcanic eruption helped in making dinos go extinct.) Real world examples include: Ash and sulfur went pretty high into Earth's atmosphere because of the Tambora eruption which in turn dimmed incoming sunlight, and lowered global temperatures by about 3°F. The Mount Pinatubo eruption in the Philippines in 1991 cooled the planet by about 1°F.