The internal energy after heating in terms of U is 100U.
The given parameters;
initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰CAssuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.
[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]
Thus, we can conclude that the internal energy after heating in terms of U is 100U.
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 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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i’m confused on this worksheet can someone help
Answer:
well for C the horse is at rest at 10 to 15
D is 0 to 10
E is 15 to 25
Explanation:
If you search up examples it will help with this worksheet.
If I am work then I appologize
Question below...............
Answer:
friction force
Explanation:
force of friction is opposite to the force applied it resist the motion
what do i do for #17
Answer:
whats the question...............
Explanation:
During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the hanging masses and other forces present is 0.91 J. At the initial time of observation the cart moves with speed 0.61 m/s. Determine the speed at the second point of observation. The mass of the cart is 0.80 Kg.
Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
The speed of the cart at the second point of observation is equal to 1.62 m/s.
What is the work done?Work can be demonstrated as the energy utilized when a force is exerted to make an object move through a particular displacement. Work done by this force is calculated from the product of the magnitude of the exerted force (F) and the distance (d) covered by the body
W= F.d
Where 'F' is the exerted force and 'd' is the displacement and W is work done.
Work and energy have a direct relationship with each other. Work done by an object can be expressed as a change in kinetic energy:
[tex]W =\frac{1}{2}mv_2^2- \frac{1}{2}mv_1^2[/tex]
Where m is the mass of the object, v₂ is the final velocity (m/s), and v₁ is the initial velocity (m/s).
Given, the work done by the cart, W = 0.91 J
The mass of the cart, m = 0.80 Kg
The speed of the cart at the first point, v₁ = 0.61 m/s
[tex]0.91 =\frac{1}{2}\times 0.80 \times v_2^2- \frac{1}{2}0.80 \times (0.61)^2[/tex]
0.40 v₂² = 1.06
v₂² = 2.65
v₂ = 1.62 m/s
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David and Fiona want to know about how mass affects the motion of an object. David uses three objects with masses of 2 kg, 3 kg, and 8 kg, writes out his procedure, and records his observations on the force needed to move the objects. Fiona is investigating three different objects using observation. What is the difference in the way they approached this question?
A. Only Fiona gathered evidence from her investigation.
B. David used scientific articles for evidence.
C. David's observations can be replicated by repeating his procedure.
D. Fiona's observations can be replicated by observing different objects.
Answer: the answer is D if its not i'm really sorry
If you have a light bulb connected to a circuit and you increase the current, what happens to the light bulb and the power (assuming the voltage remains the same)? A. The light bulb shines brighter and the power decreases. B. The light bulb shines brighter and the power increases. C. The light bulb gets dimmer and the power decreases. D. The light bulb gets dimmer and the power increases.
Answer:
B. The light bulb shines brighter and the power increases.
Explanation:
Power is the product of voltage and current . Increasing the current at a constant voltage will increase the power.
Increasing the power means more heat and therefore more light must be emitted.
PLEASEEEEEEEE HELPPPPPPP
Answer: 62.25
Explanation: F = ma
F = 7.5 * 8.3
F = 62.25
Thus, the answer is 62.25 Newtons.
Would appreciate brainliest <3
HELP ASAPPPPPPPPPPPPPP
Answer:
F=15N
Explanation:
F=m.a
m=1500g ÷1,000 = 1.5kg
a= 10m/s/s
F =1.5 × 10
F = 15 Newton
A and E are wrong i think please solve this thank you.
Answer:
I think number 3.
Explanation:
9
are things that you can achieve quickly.
O A.
Dreams
OB.
Long-term goals
O c.
Short-term goals
OD.
Plans
Answer:
d
Explanation:
A dreams
.......wkkwkwkwkwkwnwnsksk
Answer:
C. short term goal
Explanation:
took a midterm quiz on canvas, Hope this helps <3
A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun
(1 point)
O a curved line increasing from left to right
O a straight line with negative slope
O a straight line with positive slope
O a curved line decreasing from left to right
Can you please do this for me I’ll do 75 points
Answer:
Answer below, hope this helps!
Explanation:
1. Chemical - e
2.Light - b
3. Sound - g
4. Mechanical - f
5. Kinetic - a
6. Potential - d
7. Energy - j
8. Thermal - c
9. Nuclear - i
10. Electrical - h
Answer:
Chemical - e
Light - b
Sound - g
Mechanical - f
Kinetic - a
Potential - d
Energy - j
Thermal - c
Nuclear - i
Electrical - h
Explanation:
5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.
Answer:
1.904
Explanation:
F= ma
8 = 4.2 a
a = 8/4.2
a = 1.904
How is aquatic biome classification different from terrestrial biome classification?
Answer: Terrestrial biomes are based on land, while aquatic biomes include both ocean and freshwater biomes.
Explanation:
Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?
Answer:
215mph
Explanation:
multiply 112milws by 2 and multiple that by seven, then divide it by two
i need help with these please
Answer:
Explanation:
A. A small, closed chamber of gas is heated. When the gas in the
chamber expands, it does 5 J of work on a piston. The gas has
an initial energy of 8 J and a final energy of 30 J. Considering
the equation for the first law of thermodynamics (AU = Q + W.
does the work done in this scenario have a positive or negative
value? Explain.
B. How much heat is added to the gas in the chamber?
A.
The work has a negative value.
Since the gas in the chamber expands, it increases in volume and does positive work on the piston since the change in volume is positive.
Work, W = pΔV. Since the gas expands, ΔV > 0. So, W > 0. Thus Work done by the gas is positive.
Since in ΔU = Q + W, W here is work done by the surroundings, W is negative since it is the opposite of the work done on the surroundings by the gas..
So, the work has a negative value.
B.
The heat added to the gas chamber is 17 J
From the first law of thermodynamics, ΔU = Q + W where ΔU = internal energy change = U₂ - U₁ where U₁ = initial energy = 8 J and U₂ = final energy = 30 J, Q = heat added to the gas chamber and W = work on piston = - 5 J
ΔU = Q + W
U₂ - U₁ = Q + W
30 J - 8 J = Q + (-5 J)
22 J = Q - 5 J
Q = 22 J + 5 J
Q = 27 J
The heat added to the gas chamber is 27 J.
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SOMEONE HELP! If I don't get this done i will fail, I have been trying for hours.
p.s. both questions have no answer choices.
1. If a baseball has a mass of 0.075kg and is traveling with an acceleration of 25 m/s2. What is the force that the baseball will hit the bat?
````````````````````````````````````````
2. Given the force diagram of the basketball, what can you tell about the basketball motion? (Diagram below)
Thanks in advance!!
Answer:
1. The baseball will hit the bat at a force of 1.875 Newtons (N)
2. The basketball is being pushed to roll to the right of its original position with an applied rightward force. There is a force of friction to the left, a normal force of the floor acting upwards on the ball, and a gravity force of the Earth acting downwards on the ball.
Explanation:
1. To calculate force, you just have to multiply mass (m) and acceleration (a). F = m * a --> F = 0.075kg * 25 m/s^2 = 1.875 N.
2. Use your knowledge of different types of forces and free-body diagrams to come to this conclusion.
In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s?
Answer:
Explanation:
What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.
horizontal velocity component
vx = 55.0/2.50 = 22 m/s
vertical initial velocity component
(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²
vy₀ = 13.0625 m/s
θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained by the hammer head, how much will its temperature increase
The increase in temperature of the metal hammer is 0.028 ⁰C.
The given parameters:
mass of the metal hammer, m = 1.0 kgspeed of the hammer, v = 5.0 m/sspecific heat capacity of iron, 450 J/kg⁰CThe increase in temperature of the metal hammer is calculated as follows;
[tex]Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}[/tex]
where;
c is the specific heat capacity of the metal hammer
Assuming the metal hammer is iron, c = 450 J/kg⁰C
[tex]\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C[/tex]
Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.
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Identify the direction of the net force acting on the object. Explain your reasoning
Answer:
See Below
Explanation:
I am not sure what you exact question is to know what direction it is but here is how you do it.
The direction of the net force is the direction of the largest force.
for example if you were to push a box forward with 100 newton's of force and someone pushed at the same time 50 newton's backwards on the box, the box would move forwards because the was a greater force on the box in a forward direction. hope this helps
When measuring the critical angle, in which medium do we need the refracted ray to be (air or glass)
Answer:
N1 sin theta1 = N2 sin theta2 Snell's Law
For the refracted ray to be reflected then sin theta2 = 1
N1 sin theta 1 = N2
Also N2 must be less than N1 for complete reflection
sin theta1 = N2/N1
If you are considering air and glass then N2 = 1 (for air)
sin theta1 = 1 / N2 where N2 must be for glass in this case
builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
This drawing of the field lines between these two charged plates is correct.
True or false
Answer:
True
Explanation:
Since positive and negative forces are both opposites of each other they attract. But if the forces are the same where both are positive or negative thye repel do to being the same charges in that case the picture would show lines moving in opposite directions.
You throw an 8.75 g coin straight up into the air. The coin travels a distance of 337 cm upward. What was the initial speed of the coin when you released it
Answer:
hi
Explanation:
i like saying hi
Short, difficult activities that push your body are called
A.
aerobic activity
B.
anaerobic activity
C.
muscle strength
D.
exercise
Answer:
B. is your answer
Explanation:
Have a good day
Sincerly, lipor
Short, difficult activities that push your body are called anaerobic activities. Option b is correct
They are high-intensity exercises that do not require oxygen and primarily use stored energy sources within the muscles. Examples include weightlifting or sprinting.
Short, difficult activities that push your body are called anaerobic activity. Anaerobic activities are high-intensity exercises that do not require oxygen and primarily use stored energy sources within the muscles, such as weightlifting or sprinting. These activities help build muscle strength and improve power and speed.
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Which would most likely form a homogenous mixture?
Answer:
B) a pinch of sugar mixed with cup of water
Answer:
A pinch of sugar mixed with a cup of water
Explanation:
A homogeneous mixtures have a uniform appearance where the parts are pretty evenly spaced out throughout the mixture. Picking out the individual pieces should be hard, ,which is why sugar in water is the best choice.
A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is attached to the side of the tank, 0.580 m above the ground. If the valve is opened, at what speed (in m/s) will water come out of the pipe
h =(3.7 - .58)m = 3.12m
Now put PE into KE and we have to use the formula:
√2gh (g = gravity and h = height) therefor:
√2 x 9.8 x 3.12
= 7.82m/s
I hope this helps!
A wooden barrel full of water has a flat circular top of radius 25.0 cm with a small hole in it. A tube of height 8.00 m and inner radius 0.582 cm is suspended above the barrel with its lower end inserted snugly in the hole. Water is poured into the upper end of the tube until it is full. The density of water is 1.00 × 103 kg/m3.
What is the force with which the water in the barrel pushes up on the top of the barrel?
Answer:
that is all i know
Explanation:
radius= 25.0cm
height= 8m
inner radius= 0.582cm
density= 1.00 × 103kgm³f= m× a