In an elastic collision, ________. A. an individual molecule in the collision never loses energy B. the molecules involved in the collision move in a circular motion C. the total energy of all molecules in the collision remains constant D. the molecules involved in the collision are held to each other by strong intermolecular interactions
Answer:
Im pretty sure its B im very sorry if its wrong.
3
How does the electrical conductivity of metals
and metalloids change with an increase in
temperature?
Answer:
In metals there are free electrons at normal temperature so when we increase temperature it resistivity gets increases,so conductivity decreases,while in semiconductor the electrons are not free so when we increase the temperature the covalent bonds begin to break and the electron becomes free so conductivity get.
Explanation:
A strong electromagnet produces a uniform magnetic field of 1.60 T and a cross-sectional area of 0.200 m2. If we place a coil having 240 turns and a total resistance of 21.0 around the electromagnet and then we then smoothly reduced the current in the electromagnet until it reaches zero in 20.0 ms, what is the current induced in the coil
Answer:
the magnitude of the induced current is 182.86 A.
Explanation:
Given;
number of turns, N = 240 turns
cross sectional area of the loop, A = 0.2 m²
uniform magnetic field strength, B = 1.6 T
resistance of the loop, R = 21 ohms
time, Δt = 20.0 ms
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA B}{t} \\\\emf = \frac{240 \times 0.2 \times 1.6}{20 \times 10^{-3}} \\\\emf = 3,840\ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{3840}{21} \\\\I= 182.86 \ A[/tex]
Therefore, the magnitude of the induced current is 182.86 A.
an object is spun around in a circle of radius 2.0m with a period of 10.0s. what’s it’s velocity ?
Answer:
1.26 m/s
Explanation:
v=(2*pi*r)/T
v=(2*pi*2)/10
The object is spun around in a circle of radius 2.0m with a period of 10.0s. Its velocity be 1.25 m/s.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given parameters:
Radius of the circle = 2.0 m.
Period of spinning of the object = 10.0 s.
So, angular velocity of the object be = 2π/10 radian/second.
So, magnitude of velocity be = radius × angular velocity
= 2.0 × 2π/10 m/s
= 1.25 m/s.
Hence, the magnitude of velocity of the object be 1.25 m/s.
Learn more about velocity here:
https://brainly.com/question/18084516
#SPJ2
Select four of the following that would increase the magnetic field of an electromagnet
Answer:
The correct answers are: A, C, D, E
Explanation:
The magnetic field is a solenoid is given by
B = μ₀ [tex]\frac{N}{L}[/tex] I
where N is the number of turns, I the current and L length of the solenoid.
Using this equation let's examine the different responses to permute increasing the magnetic field
A) True. a thicker wire decreases the resistance and the current can increase the system.
B) False. If there is no voltage source there is no current in the system
C) True. the field is proportional to the number of turns
D) True. the magnetic moments of the core align with the field increasing its value
E) True. When the loops are closer together, more of them can fit per unit length
F) False. If the wire is shorter the number of turns decreases.
The correct answers are: A, C, D, E
1. Define force and give its Sl unit
Answer:
Force is an external agency that changes or tends to change the state of body from rest to motion or motion to rest.
The SI unit of force is newton(N)
Which of the following is true for the entropy of the universe?
A It is always decreasing.
B It is always increasing.
C It is always negative.
D It is always a constant.
Answer:
B It is always increasing.
Explanation:
In Physics, entropy can be defined as the tendency or ability of a substance to reach maximum disorder i.e to be randomly distributed.
This ultimately implies that, entropy is a thermodynamic quantity that measures the degree of maximum disorder or randomness of a system.
The S.I unit used for the measurement of the degree of maximum order or randomness of a system is Joules per Kelvin (JK¯¹). An example of entropy is the mixing of ideal gases.
Generally, the entropy in an irreversible process always increases and as such the change in entropy has a positive value.
Hence, the entropy of the universe is always increasing because its energy flow is considered to be in a downward direction rather than upward i.e from a hot region to a cold region; making the energy to be evenly distributed.
A students walks at a rate of 4 miles per hour to school. If she leaves her
house at 7:40am how long will it take her to travel 2 miles?
Answer:
30 minutes or 1/2 hour
she'll get there at 8:10am but that's not important
Explanation:
u can divide 4mph by two to find how long it would take her to travel 2 miles
she travels at 2 miles per 1/2 hour
hope this helps chu <3
A camper hears her echo off a
hillside 1.52 s after yelling. How
far is she from the hill?
Answer:
261
Explanation:
Acellus
An object, accelerating from rest at a constant rate, travels over 28 m in 11 s. What is its final velocity?
1.81 m/s
3.20 m/s
5.09 m/s
0.00 m/s
none of the above
Answer:
answer is 3
Explanation:
by using s= [(v+ u)/2] x t
28= (v+ 0)/2 x 11
v= 5.09 ms^-1
Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the fist moves through an arc length of 100 cm. What is the average speed of the fist during the hook
Answer:
average speed of the fist during the hook = 15 m/s or 1500 cm/s
Explanation:
We are given;
Speed of shoulder contraction, v_s = 75 cm/s = 0.75 m/s
Distance moved through the arc length by shoulder, d_s = 5 cm = 0.05 m
Distance moved by the fist, d_f = 100 cm = 1 m
Now, we are to find the average speed of the fist during the hook; v_f
Thus can be gotten from proportion;.
d_f/d_s = v_f/ v_s
Making V_f the subject, we have;
v_f = (d_f × v_s)/d_s
Thus;
v_f = (1 × 0.75)/0.05
v_f = 0.75/0.05
v_f = 15 m/s
Which of these statements best explains why the atmosphere of today was partly due to the interactions of spheres in the past?
Bacteria used nutrients in the soil for volcanic eruptions.
Chemicals released from water formed the atmosphere.
The molten Earth released hydrogen and helium into the atmosphere.
The increase in fertility of soil around the volcano helped produce rain.
Answer:
The answer is: Plants and animals exchange carbon dioxide and oxygen with the atmosphere.
Explanation:
Got the answer right.
3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.
A. Which bullet has more kinetic energy?
B. When you double the mass, what happens to the kinetic energy?
Answer:
A. The bullet with 0.006kg has more energy
B. When the mass is doubled the kinetic energy increases
Explanation:
Kinetic energy increases when mass increases
kinetic energy increases when velocity increases
Write an equation to help Stacy find the gravitational force on an object if she knows the mass. In the equation, let W represent gravitational force, m represent mass, and g represent the ratio you found in part D. Test your equation using a set of values from the table to be sure it works.
Answer:
The last column of the table gives this relationship:
w/m = g
Rewrite the equation to solve for W: W = m × g.
For one washer, the table shows a mass of 0.6 kilograms, a force of 5.9 newtons, and an acceleration due to gravity of 9.8 N/kg. The equation works correctly for these values and for the other values in the table:
W = m × g = 0.6 kg × 9.8 N/kg 5.9 N.
Explanation:
This is the answer on Edmentum. :)
When measuring espresso for a drink, which instrument would give the
greatest precision?
How many mL is an espresso?
One shot of espresso is generally about 30–50 ml (1–1.75 oz), and contains about 63 mg of caffeine (3). Important point: The “golden ratio” for espresso is this: a single shot is 30 to 44 mL (1 to 1.5 ounces) of water and 7 grams of coffee
Why are Big cinema hall are carpeted and their walls are made of some rough matenals.
A train with proper length L has clocks at the front and back. A photon is fired from the front to the back. Working in the train frame, we can easily say that if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c. Now consider this setup in the ground frame, where the train travels by at speed v. Rederive the above frame-independent result (namely, if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c) by working only in the ground frame.
Explanation:
In train's rest frame, the speed of photon is [tex]c[/tex] and the proper length of the train is [tex]L[/tex]. The time taken by the photon to cross the train is [tex]t=\frac{L}{c}[/tex]
In ground frame, the speed of the photon is given as follows:
[tex]v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}[/tex]
[tex]=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c[/tex]
The speed of light or photon remains same in every frame of reference.
Now, the speed of train is very less as compared to the speed of photon so that [tex]v<c[/tex] So that, [tex]\frac{v}{c} \ll 1[/tex]
The length contraction in the ground frame is given as follows:
[tex]L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}[/tex]
[tex]=L[/tex]
Time taken by the photon to travel the length of the train in ground frame is .
if mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion
Answer:
See explanations below
Explanation:
According to Newtons second law of motion
F = mass * acceleration
F = ma
If mass of an object is decreased to half, then m₂ = 1/2 m
If acting force is reduced by quarter, then F₂ = 3/4 F
F₂ = m₂a₂
3/4F = 1/2m a₂
Divide both expressions
(3/4F)/F = (1/2m)a₂/ma
3/4 = 1/2a₂/a
3/4 = a₂/2a
4a₂ = 6a
2a₂ = 3a
a₂ = 3/2 a
Hence the acceleration of its motion will be one and a half of its original acceleration.
What type of fault is shown in the image below?
In this fault, the hanging wall and footwall are side by side and moving in opposite directions.
Public Domain
Lateral fault
Normal fault
Reverse fault
Thrust fault
Answer:
this is a lateral fault
Explanation:
In a test of an energy-absorbing bumper, a 2800-lb car is driven into a barrier at 5 mi/h. The duration of the impact is 0.4 seconds. When the car rebounds from the barrier [in the opposite direction], the magnitude of its velocity is 1.5 mi/h. Use the principle of impulse and momentum to determine the magnitude of the average horizontal force (lb) exerted on the car during the impact.
Answer:
F = 2074.13 lb
Explanation:
Given that,
Mass of car, m = 2800 lb = 1270.059 kg
Initial speed, u = 5 mi/h = 2.2352 m/s
Final speed, v = - 1.5 mi/h = -0.67056 m/s (in opposite direction)
Time, t = 0.4 s
We need to find the magnitude of the average horizontal force (lb) exerted on the car during the impact. It can be calculated as :
[tex]F=m\times \dfrac{v-u}{t}\\\\F=1270.059\times \dfrac{-0.67056 -2.2352 }{0.4}\\\\F=9226.21\ N[/tex]
or
F = -2074.13 lb
So, the required force is 2074.13 lb.
What factors affect the speed of a wave? Check all that apply.
the amplitude of the wave
the energy of the wave
the temperature of the medium
the type of wave
the type of medium
Answer:
the amplitude of the wave
the energy of the wave
the type of wave
the type of medium
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
A 50kg crate is being push on a horizontal floor at constant velocity. Given that the coefficient of kenitic friction between crate and floor is hk=0.1 . What is the push force?
Answer:
F = 49 N
Explanation:
For this exercise we must use Newton's second law. Let's set a reference frame with the x axis parallel to the floor.
As they indicate that the box is going at constant speed, its acceleration is zero
Y axis y
N-W = 0
N = mg
X axis
F-fr = 0
F = fr
the friction force has the formula
fr = μ N
fr = μ mg
we substitute
F = μ m g
let's calculate
F = 0.1 50 9.8
F = 49 N
SIMPLE HARMONIC MOTION 1.0 Objective To study simple harmonic motion by observing the motion of a simple pendulum. 2.0 Simple harmonic motion Oscillatory motion is extremely common in nature. Examples include waves (water, sound, earthquake, etc.) and vibrations produced by musical instruments. If the oscillation is characterized by a constant frequency and amplitude (if the motion reproduces itself in a fixed time period T), then the motion is said to be "harmonic." If the oscillation can be described as a sinusoidal function of time and position, the motion is said to be "simple harmonic." Simple harmonic motion (SHM) occurs when for every applied force or torque, there is a restoring force or torque which is proportional to the displacement of the system from its equilibrium position. 2.0.1 Name two real-world examples of harmonic motion. (Hint: Southern California is known for what sort of natural disasters?)
Answer:
the waves in the sea, leaves of the trees, cables in the bridges, pendulum clock
Explanation:
In nature there are many examples of simple harmonic motion, for example.
* The movement of the waves in the sea is an oscillation movement up and down
* The movement of the leaves of the trees when a wind blows and then stops, but the leaf and branches are oscillating
* The movement of the cables in the bridges, especially in the suspension bridges
* The movement of a pendulum clock
if the water measures -5 feet at low tide and 3ft at high tide what is the tidal range
Answer:
8 feet
................
Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second on. If the source temperature of the first engine is 1300 K and the sink temperature of the second engine is 300 K and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is
Answer:
the temperature of the intermediate reservoir is 624.5 K
Explanation:
Given the data in the question
The two Carnot heat engines are operating in series;
[ T[tex]_H[/tex] ]
↓
((1)) ⇒ W[tex]_{out[/tex]
↓
[ T[tex]_M[/tex] ]
↓
((2)) ⇒ W[tex]_{out[/tex]
[ T[tex]_L[/tex] ]
The maximum possible efficiency for any heat engine is the Carnot efficiency;
η[tex]_{rev[/tex] = 1 - [tex]\frac{T_L}{T_H}[/tex]
the thermal efficiencies if both engines are the same will be;
η[tex]_A[/tex] = η[tex]_B[/tex]
1 - [tex]\frac{T_M}{T_H}[/tex] = 1 - [tex]\frac{T_L}{T_M}[/tex]
1 - 1 - [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]
- [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]
[tex]\frac{T_M}{T_H}[/tex] = [tex]\frac{T_L}{T_M}[/tex]
T[tex]_M[/tex]² = T[tex]_L[/tex] × T[tex]_H[/tex]
T[tex]_M[/tex] = √(T[tex]_L[/tex] × T[tex]_H[/tex])
source temperature of the first engine T[tex]_H[/tex] = 1300 K
sink temperature of the second engine T[tex]_L[/tex] = 300 K
we substitute
T[tex]_M[/tex] = √(300 × 1300)
T[tex]_M[/tex] = √390000
T[tex]_M[/tex] = 624.4998 K ≈ 624.5 K
Therefore, the temperature of the intermediate reservoir is 624.5 K
Calculate the momentum of a 10 kg bowling ball rolling at 2m/s towards north.
Answer:
momentum=mass x velocity= 10 x 2 = 20kgm/s
16–2. Just after the fan is turned on, the motor gives the
blade an angular acceleration a = (202-0.6) rad/s'. where :
is in seconds. Determine the speed of the tip P of one of the
blades when i 3 s. How many revolutions has the blade
turned in 3 s? When / O the blade is at rest.
Answer:
P.S My answer may not render so let me know if it doesn't
Explanation:
When we add or remove energy from a substance, what kind of changes can we observe? Can they happen at the same time ?
Answer:
When you add energy to a substance?
One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules. The temperature at which a substance goes from a solid to a liquid is it melting point.
When you remove energy from a substance?
When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.
Explanation:
The addition of energy increases the kinetic energy of the particles, which reduces the intermolecular forces between the particles. Freezing occurs when a liquid becomes a solid and energy is released.
Atoms lose energy when they change from solid to liquid or gas and quid to gas.