c. a natural process
It is a natural process
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Where does the extra energy go?
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Then the extra energy goes in overcoming the gravitational acceleration and heating up of body etc
Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer day in Maryland. Use $0.17 per kWh as your energy cost.
Answer:
The cost of energy is $ 0.34.
Explanation:
The energy is the capacity to do work.
The energy is a scalar quantity and its SI unit is Joule.
The commercial unit of energy is kWh.
Cost of 1 kWh energy = $ 0.17
energy loss by standard window is 2 kWh .
So, the cost of lost of energy is
Cost = $ 0.17 x 2 = $ 0.34
A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42
m/s. What is the initial kinetic energy of the train?
Answer:
kinetic energy of the train = 2,910.6 x 10⁷ joule
Explanation:
Given:
Mass of train = 3.3 x 10⁷ kg
Speed of train = 42 m/s
Find:
kinetic energy of the train
Computation:
kinetic energy = (1/2)(m)(v²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)
kinetic energy of the train = (3.3 x 10⁷)(882)
kinetic energy of the train = 2,910.6 x 10⁷ joule
Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Explanation:
Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]
Speed = 42 m/s
Kinetic energy is the energy acquired by an object due to its motion.
Formula to calculate kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
m = mass of object
v = speed of object
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]
Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Walking at a brisk pace, you cover 10 m in 5.0 s . How many seconds will you need to cover 50 m ?
pls solve this:
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Answer: The frequency of a pendulum is [tex]0.2 s^{-1}[/tex].
Explanation:
Time period is defined as the time required to produce complete wave.
As we know that the frequency and time are inversely proportional to each other.
That means,
[tex]\nu =\frac{1}{T}[/tex]
Where,
[tex]\nu[/tex] is frequency of pendulum
T is time period
Given:
Time period = 5 seconds
Now putting all the given values in the above formula, we get the frequency of the pendulum.
[tex]\nu =\frac{1}{T}\\\\\nu =\frac{1}{5s}\\\\\nu =0.2s^{-1}[/tex]
Therefore, the frequency of a pendulum is [tex]0.2 s^{-1}[/tex].
An athlete training for an event does 6.53 104 J of work during a workout and gives off 5.97 105 J of heat. Consider the athlete to be like a heat engine. (a) Determine the magnitude of the change in internal energy of the athlete. J (b) What is the efficiency of the athlete
Answer:
(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J
(b) the efficiency of the athlete is 10.94 %
Explanation:
Given;
work done by the athlete (system), W = 6.53 x 10⁴ J
the heat given off by the athlete (system), Q = 5.97 x 10⁵ J
The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.
Q← ⊕ →W
The work, W, points away from the system since the system does the work
The heat, Q, points away from the system since heat is given off
Apply first law of thermodynamic;
ΔU = Q + W
where;
q is the heat flowing into or out of the system
(+q if the heat is flowing into the system
(-q if the heat is leaving the system
w is the work done by or on the system
(+w if the work is done on the system by the surrounding
(-w if the work is done by the system to the surrounding
Thus, from the above explanation, the change in internal energy of the system is calculated as;
ΔU = -Q - W
ΔU = - 5.97 x 10⁵ J - 6.53 x 10⁴ J
ΔU = -6.623 x 10⁵ J
The magnitude of the change in internal energy = 6.623 x 10⁵ J
(b) the efficiency of the athlete;
[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]
different between pressure and force
Force is mass into acceleration
and pressure is force applied per unit area.
How is a continuous spectra is formed?
Answer:
Hello There!!
Explanation:
They are produced by the photodissociation of negatively charged hydrogen ions (H−).
hope this helps,have a great day!!
~Pinky~
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 2I0?
A. f/4
B. f/2
C. 3f/2
D.) 2f
Answer:
Option D
Explanation:
From the question we are told that:
The attractive magnetic force per unit length as
[tex]f = F/L[/tex]
Separation Distance [tex]x=2d[/tex]
Generally the equation for Magnetic force between two current carrying wire is mathematically given by
[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]
[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]
Where
[tex]x=2r[/tex]
And
[tex]I_1=I_2=>2I[/tex]
Then
[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]
[tex]\frac{F}{\triangle l}=>2f[/tex]
Therefore s the force per unit length between the wires if their separation is 2d
[tex]\frac{F}{\triangle l}=>2f[/tex]
Option D
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
What are impact and non-impact printers?
Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.
In Non-Impact printers, no mechanical moving component is used.
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If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m/s2, what is the mass of the box
Answer:
The correct solution is "14.6875 kg".
Explanation:
Given values:
Force,
F = 47.0 N
Acceleration,
a = 3.20 m/s²
Now,
⇒ [tex]Force=Mass\times Acceleration[/tex]
or,
⇒ [tex]F=ma[/tex]
⇒ [tex]47.0=m\times 3.20[/tex]
⇒ [tex]m=\frac{47.0}{3.20}[/tex]
⇒ [tex]=14.6875 \ kg[/tex]
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
is anyone online??just asking
Answer:
me...:(
Explanation:
Answer:
hello I'm online here thanks for the points (◔‿◔)
g You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
what's the difference between mass and inertia in a tabular form
Answer:
to be honest i dont know
Explanation:
^^
If a second ball were dropped from rest from height ymax, how long would it take to reach the ground
Answer:
[tex](b)\ t_1 - t_0[/tex]
[tex](d)\ t_2 - t_1[/tex]
[tex](e)\ \frac{t_2 - t_0}{2}[/tex]
Explanation:
Given
See attachment for complete question
Required
How long to reach the ground from the maximum height
First, calculate the time of flight (T)
[tex]T =t_2 - t_0[/tex]
The time taken (t) from maximum height to the ground is:
[tex]t = \frac{1}{2}T[/tex]
So, we have:
[tex]t = \frac{t_2 - t_0}{2}[/tex]
Another representation is:
At ymax, the time is: t1
On the ground, the time is t2
The difference between these times is the time taken.
So;
[tex]t = t_2 - t_1[/tex]
Since air resistance is to be ignored, then
[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height
At which location would a bowling ball have the greatest weight?
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/sm/s. Part A What is the speed of the putty just before it strikes the ceiling
Answer: [tex]4.65\ m/s[/tex]
Explanation:
Given
Distance putty has to travel is 3.5 m
The initial speed of putty is 9.50 m/s
Using equation of motion to determine the velocity of putty just before it hits ceiling
[tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]
So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
A good soccer player can kick the ball up to 25 m/sec. A soccer ball has a mass of 800 grams (0.8 kg). What force must a goalie exert on the ball to bring it to rest in 0.1 sec?
Answer:
200 N
Explanation:
Applying,
The force a golie must exert on the ball is,
F = ma...................... Equation 1
Where m = mass of the ball, a = acceleration of the ball.
But,
a = Δv/t............... Equation 2
Where Δv = change in velocity, t = time.
Substitute equation 2 into equation 1
F = m(Δv/t)............... Equation 3
From the question,
Given: m = 0.8 g, t = 0.1 s, Δv = 25 m/s
Substitute these values into equation 3
F = 0.8×25/0.1
F = 200 N
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object. For an object to change its state of motion, a force must be applied to it.
A. Velocity
B. Inertia
C. Distance
D. Area
Answer:
B) Inertia is the resistance of any physical object
How much work can a motor with a power output of 0.70 hp do in 2 s?
Answer:
the work done by the motor is 1,044 J.
Explanation:
Given;
the output power of the motor, P = 0.7 hp
duration of the work, t = 2 s
The relationship between horse-power and watt is given as;
1 hp = 745.7 W
0.7 hp = ?
0.7 hp = 522 W = 522 J/s
The work done by the motor is calculated as;
W = Power x time
W = 522 J/s x 2 s
W = 1,044 J
Therefore, the work done by the motor is 1,044 J.
The pair of forces described by Newton third law must be
Answer:
The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs
The pair of forces described by Newton third law must be in opposite direction.
What is Newton's third law of motion ?Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.
The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.
When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.
A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.
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А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
Complete question:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.
Answer:
The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Explanation:
Given;
power of the flux produced, P = 1 W = 1 J/s
energy per photon, E = 78 keV
Convert the energy per photon to J
E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon
let the number of photons = n
n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s
[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]
Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s. Calculate his:
a. acceleration in his direction of motion
b. acceleration opposite to his direction of motion.
Answer: [tex]56.4\ m/s^2, 201.42\ m/s^2[/tex]
Explanation:
Given
Rocket attain a velocity of [tex]v=282\ m/s[/tex] in a time period of [tex]t=5\ s[/tex]
It was brought jarringly back to rest in only [tex]t'=1.4\ s[/tex]
Acceleration is the change in velocity of the object over a period of time
(a) Acceleration in his direction of motion
[tex]\Rightarrow a=\dfrac{v-0}{t}\\\\\Rightarrow a=\dfrac{282}{5}\\\\\Rightarrow a=56.4\ m/s^2[/tex]
(b) acceleration opposite to his direction of motion i.e. deceleration is
[tex]\Rightarrow a_d=\dfrac{0-v}{t'}\\\\\Rightarrow a_d=\dfrac{-282}{1.4}\\\\\Rightarrow a_d=-201.42 \ m/s^2\\\Rightarrow a_d=201.42\ \text{decelration}[/tex]
What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per second (m/s) and assume acceleration from gravity is 10 m/s2.
speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
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