To pull a 50 kg crate across a horizontal frictionless floor, a worker applies a force of 210 N, directed 20° above the horizontal. As the crate moves 3.0 m, what work is done on the crate by (a) the worker's force, (b) the gravitational force on the crate, and (c) the normal force on the crate from the floor? (d) What is the total work done on the crate?
Answers
The work done on the crate by the worker's force is 592 J.
Given that,
Mass of the crate, m = 50kg
Force applied by the worker, F = 210N
Angle to the horizontal, θ = 20°,
Distance moved by crate, d=3m
A force causes an object with mass to change its velocity, therefore, the acceleration. acceleration caused by a force is always in the direction of force applied.
Also, we know that work done is always given as the product of force and displacement in the direction of the force.
(a.) work done on the crate by the worker's force,
As the force is been applied at an angle directed 20° above the horizontal.
Therefore, the force is causing the crate to move in the horizontal direction, thus, only the horizontal component of the force is taken.
(b) work done on the crate by the gravitational force,
As the gravitational force is always applied perpendicular to the horizontal.
Therefore, the gravitational force is not causing the crate to displace anywhere, thus, the displacement is zero.
Hence, work done on the crate by the gravitational force is 0 J.
(c.) work done on the crate by the normal,
Normal is the contact force that is perpendicular to the surface with which the object has contact.
As the normal force did not cause the crate to displace upward by any distance.
(d.) The total work done on the crate,
As discussed above the only work done on the crate is by the worker.
Therefore, The total work done on the crate is 592 J.
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Answer:
Explanation:
Given that,
A crate of mass M = 50kg
The crate is pulled along an horizontal floor by a string at an angle
Force pulling the crate
F = 210N
Angle θ = 20° to the horizontal
Distance moved by crate d=3m
A. Work done by force
W = FdCosθ
W = 210 × 3 × Cos20
W = 592J
B. Work done by Gravitational force?
Work is define as the dot product of force and displacement in the direction of the force.
Since the gravitational force does not cause the crate to move any distance downward
Then,
W(gravity) = mg×d
distance d=0
W(gravity) = 0 J
C. Work done by normal?
Since, the normal force did not cause the crate to move upward by any distance
Then,
W(normal) = 0J
D. The total workdone?
The only workdone on the crate is by the person
Then, the total workdone is the workdone by the person
W = 592 J
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Answers
Answer:
Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that
here we will have
so we have
so we have
Final answer:
The concept of torques and equilibrium is used to calculate the pulling force on the larger flywheel, which is found to be approximately 29.55 Newtons. This force will balance the system and prevent it from rotating.
Explanation:
To solve this problem, we need to understand the concept of torque and equilibrium. We know that torque (τ) is the rotational equivalent of linear force. It's calculated by the formula τ = force × radius. Thus, for the system to stay at equilibrium (not rotate), the torques need to balance each other out.
On the smaller flywheel, the torque τ₁ is given by the pulling force (F₁ = 50 N) and the radius (r₁ = 13 cm, or 0.13 m), hence τ₁ = F₁ × r₁ = 50 N x 0.13 m = 6.5 N.m.
In order for the system to stay at equilibrium, the same amount of torque needs to be applied to the larger flywheel. We already know the radius of the larger flywheel (r₂ = 22 cm, or 0.22 m). To keep the system at equilibrium, the pulling force F₂ on the larger flywheel should be such that the torque τ₂ = τ₁ = 6.5 N.m. From the formula τ = F × r, we can solve for F₂ as follows: F₂ = τ₂ / r₂ = 6.5 N.m / 0.22 m = 29.55 N, approximately. Therefore, a pulling force of about 29.55 N should be applied to the cord connected to the larger flywheel to prevent the system from rotating.
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refraction
superposition
diffraction
Answers
Answer : diffraction
Explanation : Diffraction is the phenomenon of bending of sound waves around the small openings or obstacles.
Tierra is playing in her backyard when she hears her friend calling out to her. Tierra can hear her friend even when she can't see her. This is because the sound is spreading out from the small openings and hence diffraction of sound wave occurs.
Answers
Answer:
Explanation:
One
I've never heard of a mass defect before you asked this question. Interesting concept.
Givens
Number of neutrons =207 - 82 = 125
Number of protons = 82
The amu of a proton is 1.00728
The amu of a neutron is 1.00866
Solution
Find out the actual mass of the Lead atom
Actual mass = 82*1.00728 + 125*1.00866
Actual mass = 208.67946
The mass defect = 208.67946 - 207 = 1.67946
That is the proper way to calculate mass defect.
Apparently the way you are supposed to do it is
Mass defect = 208.67946 - 206.975897= 1.703563 which means that C is the closest answer. You will have to look at your notes to see if amu for protons and neutrons are what I used.
Two
The number of protons is the number of Uranium on the periodic table.
92 protons. 92 comes from the position on the periodic table.
You can get the same thing by Taking its relative mass (238) and subtracting the neutrons (146) which is 238 - 146 = 92
Answer:
92 protons
Explanation:
The mass number is 238, so the nucleus has 238 particles in total, including 146 neutrons. So to calculate the number of neutrons we have to subtract: 238 − 146 = 92
Answers
The pressure of the sample of gas at 273 K is equal to 0.273 atm assuming the volume is constant.
What is Gay Lussac's law?
Gay-Lussac's law can be described as that when the volume of the gas remains constant then the pressure (P) of the gas and the absolute temperature (T) in a direct relationship.
Gay Lussca's law can be described as mentioned below:
P/T = k
The pressure of gas is directly proportional to absolute temperature of the gas.
P ∝ T ( Volume of gas is constant)
or, P₁/T₁ = P₂/T₂ .................(1)
Given, the initial temperature of the gas, T₁ = 400 K
The final temperature of the gas, T₂ = 273 K
The initial pressure of the gas, P₁ = 0.40 atm
The final pressure of the given gas can be calculated from Gay Lussac's law as:
P₂ = (0.40/400) ×273
P₂ = 0.273 atm
Therefore, the pressure of this gas at 273 K is 0.273 atm.
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Answers
Answer:
(9/35) = 0.257
Explanation:
Box contains 9 new light bulbs and 6 used light bulbs, total number of bulbs = 15.
Probability of selecting two bulbs; a new light bulb and then, a used light bulb in that order = [(probability of selecting a new bulb) × (probability of selecting a used bulb from the rest)] = [(9/15) × (6/14)] = (9/35) = 0.257
Final answer:
The probability that Meredith will select a new bulb and then a used bulb in sequential order without replacement is around 26%.
Explanation:
The question asks about the probability of selecting a new light bulb and then a used light bulb from a box containing 9 new light bulbs and 6 used light bulbs. Probability events like these are solved using multiplication rules of probability that each event is independent.
First, the probability of picking a new bulb is 9/15 (total bulbs are 15). If you pick one out, you don't replace it, so there are only 14 bulbs left. Thus, the probability of picking a used bulb now is 6/14. Hence the probability of both events happening in order is the multiplication of both, i.e., (9/15)*(6/14) = 54/210 = 0.257 or around 26%.
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