Answer:
The power output is [tex]P = 360 \ W[/tex]
Explanation:
From the question we are told that
The potential difference developed is [tex]V = 450 V[/tex]
The current is [tex]I = 0.80 \ A[/tex]
The duration of the pulse is [tex]\Delta t = 1.0 ms = 1,0 *10^{-3} \ s[/tex]
Generally the power output of the eel during the pulse is mathematically represented as
[tex]P = V * I[/tex]
=> [tex]P = 450 * 0.80[/tex]
=> [tex]P = 360 \ W[/tex]
Jaeda designs two electromagnets of different strengths. The first electromagnet has an insulated copper wire wrapped around an iron nail many times before it is connected to a battery. The second electromagnet has its insulated copper wire wrapped around a similar nail a fewer number of times before it is connected to two smaller batteries as shown in the image. Jaeda tries to determine which electromagnet is stronger by bringing each near a bowl containing eight metal paper clips. She observes that both electromagnets attract all eight of the paper clips.
How can Jaeda demonstrate that each electromagnet has a different strength
A. Place a metal paper clip on a table and slowly bring each electromagnet closer to the paper clip to see which can attract the paper clip from a greater distance.
B. Determine which electromagnet has its copper wire wrapped around its iron nail a greater number of times.
C. Add plastic paper clips to the bowl and check which electromagnet can attract both kinds of paper clips.
D. Reduce the number of paper clips in the bowl and find which electromagnet attracts more paper clips.
Answer:A
Explanation:
Place a metal paper clip
g A 320-g air track cart is traveling at 1.25 m/s and a 270-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts
Answer:
1.287m/s
Explanation:
Using the law of conservation of mass expressed as;
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the objects
u1 and u2 are the velocities
v is the speed of the center of mas of the two carts
Given
m1 = 320g = 0.32kg
u1 = 1.25m/s
m2 = 270g = 0.27kg
u2 = 1.33m/s
Substitute this value into the formula and get v;
0.32(1.25)+0.27(1.33) = (0.32+0.27)v
0.4+0.3591 = 0.59v
0.7591 = 0.59v
v = 0.7591/0.59
v = 1.287m/s
hence the speed of the center of mass of the two carts is 1.287m/s
a steel is 40cm long at 20c.the coefficient of linear expansivity for steel is 0.000012.find the increase in length at 70C
Answer:
janakajnakanabjjiaj yav good day for you to everyone else to do with my family is it just doesn't want me and I am
Help me with this please !!
Pt 1_It would be harder to do experiments on or near jupiter for many reasons ! One being that jupiters top layers are known to be gas. However if we could hover over Jupiter and test its gravity, describe or sketch the shape of the position vs time graph for a dropped object.
PT2_ The acceleration due to gravity near Jupiter is 25.95m/s. If a hammer were dropped, how fast wouldit be going after falling 20.0 meters. Show your work !
Answer:
he is right he Is
Explanation:
step by step explanation
A 10.0 kg mass is attached to the end of a 2.00 m long brass rod, which has a diameter of 1.00 mm and negligible mass. The mass at the end is pulled, stretching the rod slightly, and then released. If the elastic modulus of brass is 9.10 × 1010 N/m2, then the period of the resulting oscillations is
A. 0.175 sec.
B. 0.105 sec.
C. 0.133 sec.
D. 0.145 sec.
E. 0.167 sec.
Answer:
The appropriate alternative is Option B (0.105 sec.).
Explanation:
The given values are:
Elastic modulus,
Y = 9.10 × 10¹⁰ N/m²
Mass,
m = 10.0 kg
Length of rod,
l = 2.00 m
Diameter,
d = 1.00 mm
Now,
⇒ [tex]Keq=\frac{AY}{l}= \frac{\pi D^2 Y}{4l}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{\pi \times 10^{-6}\times 9.1\times 10^{10}}{4\times 2}[/tex]
⇒ [tex]=3.574\times 10^4[/tex]
The time period will be:
⇒ [tex]T=2\pi \sqrt{\frac{m}{Keq} }[/tex]
On substituting the above values, we get
⇒ [tex]=2\pi \sqrt{\frac{10}{3.574\times 10^4}}[/tex]
⇒ [tex]=0.105 \ seconds[/tex]
The time period resulting oscillations will be 0.1005 seconds.
What is the time period of oscillation?The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
The given data in the problem is;
[tex]\rm \gamma[/tex] is the elastic modulus=9.10 × 10¹⁰ N/m²
m is the mass= 10.0 kg
l is the length of brass rod= 2.00 m
d is the diameter of 1.00 mm
The value of the equivalent stiffness will be;
[tex]\rm K_{eq}= \frac{AY}{l}\\\\ \rm K_{eq}= = \frac{\pi d^2 Y}{4l} \\\\ \rm K_{eq}=\frac{3.14 \times 10^{-6}\times 9.1 \times 10^{10} }{4\times 2 } \\\\ \rm K_{eq}= 3.574 \times 10^4[/tex]
The time period of the oscillation is given by;
[tex]\rm T = 2 \pi \sqrt{\frac{m}{k_{eq}} } \\\\ \rm T = 2 \times 3.14 \sqrt{\frac{10}{3.574 \times 10^4}[/tex]
[tex]\rm T = 0.105 \ sec[/tex]
Hence the time period resulting oscillations will be 0.1005 seconds.
To learn more about the time period of oscillation refer to the link;
https://brainly.com/question/20070798
Bru challenge with the physics.
Answer:
WHATATATATTAT
Explanation:
.
An oatmeal cookie is dropped on the floor. Is this an inelastic collision?
Answer:
It is not inelastic because the cookie and floor do not move away together as a unit.
Explanation:
You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged
that your client, Mr. Lawton, shot the victim, Mr. Cray. The detective who investigated the scene of the
crime, Mr. Dibny, found a second bullet, from a shot that missed Mr. Cray, that had embedded itself into a
chair. You arise to cross-examine the detective.
You: In what type of chair did you find the bullet?
Dinby: A wooden chair.
You: How massive was this chair?
Dinby: It had a mass of 20 kg.
You: How did the chair respond to being struck with a bullet?
Dinby: It slid across the floor.
You: How far?
Dinby: Three centimeters. The slide marks on the dusty floor are quite distinct.
You: What kind of floor was it?
Dinby: A wood floor.
You: What was the mass of the bullet you retrieved from the chair?
Dinby: Its mass was 10 g.
You: Have you tested the gun you found in Mr. Lawton's possession?
Dinby: I have.
You: What is the muzzle velocity of bullets fired from that gun?
Dinby: The muzzle velocity is 450 m/s.
With only slight hesitation, you turn confidently to the jury and proclaim, "My client's gun did not fire those
shots!"
(a) How are you going to convince the jury and judge?
(b) Choose one part of your solution and perform a sense-making analysis. Clearly state which sensemaking analysis you’ve chosen and why.
Answer:
It was not fired from the client's gun because the chair slid only 3 centimeters . If it had been fired from the client's gun the chair would slid 25.82 centimeters.
Explanation:
According to the law of conservation of momentum the momentum of the system before collision must be equal to the momentum of the system after the collision.
M1u1= m2u2
Let M1 = mass of the chair = 20kg
m2= mass of the bullet= 10g= 0.001kg
u1= velocity of the chair before collision = zero m/s
u2 = velocity of the bullet before collision = zero m/s
v1= velocity of the chair after collision = ? m/s
v2 = velocity of the bullet after collision = 450 m/s
After collision their velocities change from u1 to v1 and u2 to v2 so
M1v1= m2v2
v1= m2v2/M1
v1= 0.01 *450/ 20= 0.225 m/s
Now according to the law of conservation of energy the energy of the system before collision must be equal to the energy of the system after the collision.
The energy of the chair after the bullet is hit is
KE of the chair + KE of the bullet= 1/2 (M)(v1)²+ 1/2 m(v2)²=
1/2 ( 20) (0.225 )² + 1/2 (0.01) (450)²
= 0.50625 + 1012.5= 1013.00625 Joules
Frictional force = Coefficient of kinetic force of wood on wood ( M+m) g
= 0.2* ( 20.01) 9.8= 39.2196 N
Work done by friction = frictional force * distance
If law of conservation of energy is applied the KE must be equal to the work done
KE = W
W= f*d
KE= F*d
d = KE/f= 1013.00625/ 39.2196= 25.82 cm
The chair did not move 25.82 cm .
It only moved 3 centimeter.
Hence the bullet fired was not from the client's gun.
it opens or close the circuit
Answer:
The person above me is right i had a test a couple of days ago and thats kinda what u put and got it right!
The CERN particle accelerator is circular with a circumference of 7.0 km.
Required:
a. What is the acceleration of the protons (m=1.67×10^−27kg) that move around the accelerator at of the speed of light? (The speed of light is v=3.00×10^8m/s.)
b. What is the force on the protons?
Answer:
Explanation:
a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;
a = mv²/r
m is the mass of proton = 1.67×10^−27kg
v is the velocity = 3.00×10^8m/s
r is the radius
Since C = 2πr
7000m = 2πr
r = 7000/2π
r = 1114.08m
Substitute
a = 1.67×10^−27 (3.00×10^8)²/1114.08
a = 1.67×10^−27 * 9×10^16/1114.08
a = 15.03*10^-11/1114.08
a = 0.001346*10^-11
a = 1.346*10^-14m/s²
b) Force on the proton = mass * acceleration
Force = 1.67×10^−27kg * 1.346*10^-14
Force = 2.246*10^-41N
hence the force on the proton is 2.246*10^-41N
A block, whose mass is 0.500 kg, is attached to a spring with a force constant of 126 N/m. The block rests upon a frictionless, horizontal surface. A block labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the block is displaced by a distance A to the right of its equilibrium position. The block is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The block is then released from rest. (a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the block? N (b) At that very instant, what is the magnitude of the block's acceleration (in m/s2)? m/s2 (c) In what direction does the acceleration vector point at the instant of release?
Answer:
(a) F = 15.12 N
(b) a = 30.24 m/s²
(c) To Left
Explanation:
(a)
The magnitude of the spring force is given by Hooke's Law as follows:
F = kx
where,
F = Spring Force = ?
k = Spring Constant = 126 N/m
x = Displacement = A = 0.12 m
Therefore,
F = (126 N/m)(0.12 m)
F = 15.12 N
(b)
The magnitude of acceleration can be found by comparing the spring force with the unbalanced force formula of Newton's Second Law:
F = ma
where,
F = Spring Force = 15.12 N
m = mass of block = 0.5 kg
a = magnitude of acceleration = ?
15.12 N = 0.5 kg (a)
a = 15.12 N/0.5 kg
a = 30.24 m/s²
(c)
Since, the acceleration is always directed towards mean (equilibrium) position in periodic motion. Therefore, the direction of the acceleration at the time of release will be to left.
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.3 s. The readout on the display indicates that the average power you are producing is 92 W. What is the magnitude of the force that you exert on the handle?
Answer:
The magnitude of force exerted on the handle is 108.73 N
Explanation:
To determine the magnitude of force exerted, we will use the formula relating Power and Force.
Power is the rate at which work is done. Power can be calculated from the formula
Power = Work / Time
But, Work = Force × Distance
Hence,
Power is given by the formula
[tex]P = \frac{F \times s}{t}[/tex]
Where P is the Power
F is the force
s is the distance
and t is the time
From [tex]P = \frac{F \times s}{t}[/tex],
Then we can write that
[tex]F = \frac{P \times t}{s}[/tex]
From the question,
Distance, s = 1.1 m
Time, t = 1.3 s
Power, P = 92 W
Putting these values into the formula, we get
[tex]F = \frac{92 \times 1.3}{1.1}[/tex]
[tex]F = \frac{119.6}{1.1}[/tex]
[tex]F = 108.73N[/tex]
Hence, the magnitude of force exerted on the handle is 108.73 N.
