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sweet-ann
7 days ago
15

A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the sta

rting temperature of the cold water is 16'c? The specific heat capacity of water is 4200 J/Kg'c
The two equations you can use is in the picture.

Physics
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A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts
inna [3103]
Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

3 0
2 months ago
The spring is now compressed so that the unconstrained end moves from x=0 to x=L. Using the work integral W=∫xfxiF⃗ (x⃗ )⋅dx⃗ ,
Sav [3153]

solution:

the spring force applied by a spring with spring constant k can be expressed as

F(x)=-kx

where k acts as the spring constant

and x indicates the spring's deformation

to determine the work completed by the spring

W=\int\limits^L_0 {} \, dW

the amount of work done by the spring when moving from x=0 to x=L

W=-kx^2/2

substituting the limits x=0 and x=L

we derive the work done in terms of k and L

ANSWER

W=-kL^2/2

4 0
1 month ago
Joan and Mike's teacher show them a picture of an atom and ask them to identify it using the periodic table. Joan says the atom
ValentinkaMS [3465]
Since we can't see the image the teacher has provided, we must deduce its content based on clues from the answer choices. The image likely illustrates an atom containing 3 protons and 3 neutrons in the nucleus, along with 3 electrons orbiting it. If that is the case, Mike is correct (first choice) because the atomic number indicates the number of protons present in an atom. Since there are 3 protons, it corresponds to lithium, which has an atomic number of 3. The second choice is false, as electrons do not factor into atomic mass. The third and fourth options are also incorrect; regrettably, Joan does not grasp the concepts accurately.
3 0
1 month ago
An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time
Keith_Richards [3271]

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

3 0
3 months ago
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