Answer:
3 * 10³J/s
Explanation:
Given :
Force applied, F = 300 N
Distance, d = 30 m
Time, t = 3 seconds
Power, P = Workdone / time
Recall :
Workdone = Force * distance
Workdone = 300 N * 30 m = 9000 Nm
Workdone = 9 * 10³ J
Power = (9 * 10³ J) / 3s
Power = 3 * 10³J/s
Why must you bend forward when carrying a
heavy load on your back?
1. The gravitational force has decreased.
2. Angular momentum has decreased.
3. The center of gravity has shifted.
4. Inertia has changed.
Answer:
hi I thinks its number 3
Explanation:
hope you have a nice day
Protons, neutrons, electrons, and a nucleus are
Frequency more than 20,000 HZ
Answer:
dvhd
Explanation:
xhxjjdvcbxhjddvifidid
Answer:
The units of frequency are called hertz (Hz). Humans with normal hearing can hear sounds between 20 Hz and 20,000 Hz. Frequencies above 20,000 Hz are known as ultrasound
What does a step-up transformer do?
A. It steps up the energy.
B. It steps up the power.
C. It steps up the voltage.
D. It steps up the current.
Does latitude has an effect on weight? PLEASE HELP!
Answer:
yes
it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.
A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance from the wire of A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance from the wire of:_____.
a. 4d.
b. 16d.
c. 2d.
d. 8d.
Answer:
A. 4d
Explanation:
Let's begin with the formula for the magnetic field produced by a long wire.
[tex]B = \frac{\mu_0I}{2\pi d}[/tex]
So [tex]d=\frac{\mu_0 I}{2\pi B }[/tex]
at point d_{1} is
[tex]d_{1}=\frac{\mu_{0} i}{2 \pi B_{1}} \\ \frac{d_{1}}{d}=\frac{\frac{\mu_{0} i}{2 \pi B_{1}}}{\frac{\mu_{0} i}{2 \pi B}} \\ d_{2}=d\left(\frac{B}{B}\right) \\ =d\left(\frac{20 \mathrm{mT}}{5 \mathrm{mT}}\right) \\ =4 d[/tex]
Hence, option (A) is correct answer
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Distance travelled in one round will be equal to the circumference of the disc i.e [tex] 2 \pi r[/tex]
Radius=50cm
Circumference= [tex] 2 \times \frac{22}{7} \times 50cm=> \frac{2200}{7} cm[/tex]
If the disc rotates at speed of 100rpm that means it completes 100 rotation in a minute(60 second)
So, in 30 seconds it will complete 50 rotation.
1 rotation = [tex] \frac{2200}{7} cm [/tex]
[tex] 50 rotation=\frac{2200}{7} \times 50cm \\\\ \frac{110000}{7} cm[/tex]
Break it in decimal.
Your answer will be 15714.2 cm
7) If someone behaves against your company's code of ethics, what should you do?
A) Ignore it and mind your own business.
B) Suggest that they talk to the human resources department.
C) Talk to them about the situation.
D) Report the problem to your supervisor
Answer:
D) Report the problem to your supervisor.
Explanation:
This is probably the most efficient way get them to stop or to get them to follow the rules <3
Scroll over the answer choices to see the images. Choose all of the true statements concerning the corresponding image
The picture depicts an electric motor which turns electrical energy into mechanical energy.
The picture depicts a generator which turns electrical energy into mechanical energy
A series circuit is a good example of an electromagnet
F more wite is wound around this iron mail, the strength of the electromagnetic is increased
This image devices that the direction of the magnetic field does not depend on the direction of the current
Flow of electron |
Answer:
Explanation:
I jus did it on usatestprep
Which one of the statements below is true about mechanical waves?
They must travel in empty space.
They can travel in a vacuum.
Both sound and light are examples of mechanical waves.
They require a medium to travel through.
Answer:
D) Mechanical waves require a medium for transmission (wire, air, etc.) as opposed to electromagnetic which travel through empty space - light, radio, etc.)
Which one of the following statements is not true of free falling object
Answer:
FORCE as for my answer....
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
PROJECT: VIRTUAL LABS — CIRCUITS
Answer:
Table A
Measuring Current as a Function of Voltage with a 20 Ω Resistor
Voltage
(V)
Current: Calculated
(A)
Current: Experimental
(A)
1 0.05 0.05
5 0.25 0.25
10 0.50 0.50
20 1.00 1.00
50 2.50 2.50
Table B
Measuring Current as a Function of Resistance at 25 V
Resistance
(Ω)
Current: Calculated
(A)
Current: Experimental
(A)
10 2.50 2.50
20 1.25 1.25
100 0.25 0.25
200 0.12 0.12
Table C
Measuring Current in a Parallel Circuit
Resistor Set
(Ω)
Total
Resistance
(Ω)
Calculated
Current
(A)
Observed
Current
(A)
Observed Current
through Each Resistor
(A)
20, 20, 20 6.67 3.75 3.74 1.25, 1.25, 1.25
20, 20, 200 9.52 2.63 2.62 1.25, 1.25, 0.12
Voltage needed to raise current to 3.75 A (20, 20, 200 resistor set):
Calculated: 35.7
Observed: 36
Table D
Calculating Power of Circuit ComponentsTeacher Guide (continued)
Observed Total Current
(A)
Current through Each Bulb
(A)
Power Usage per Bulb
(W)
2.00 0.67 6.7
Explanation:
got this from the teachers guide
Determiner l'interfrange i sur le plan d'observation π distant de L de D:
Determine the interfringe i on the observation plane π distant from L from D:
1) pour D=f
if D=f
2)pour D=2f
if D=2f
Answer:
can you explain in Hindi language
because i learn hindi
What do meteorologists call the lines
that join places with the same
temperature?
A. isobars
B. isotherms
C. anisobars
D. anisotherms
Question 8: Unspooling Thread (100 points) A 110 g spool of thread with a 4.2 cm radius is held up by a peg through its center and allowed to freely rotate. Assume the thread is ideal (i.e., it does not stretch or slip, and its mass is negligibly small). A 160 g needle is tied to the loose end of the thread. The needle is dropped, and it accelerates to the floor as the thread unwinds. Find the tension in the thread and the magnitude of the acceleration of the needle as it falls.
Answer:
a = 7.29 m / s², T = 0.40 N
Explanation:
To solve this exercise we must apply Newton's second law to each body
The needle
W -T = m a
mg - T = ma
The spool, which we will approach by a cylinder
Σ τ = I α
T R = I α
the moment of inertia of a cylinder with an axis through its center is
I = ½ M R²
angular and linear variables are related
a = α R
α = a / R
we substitute
T R = (½ M R²) a / R
T = ½ M a
we write our system of equations together
mg - T = m a
T = ½ M a
we solve
m g = (m + ½ M) a
a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]
let's calculate
a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]
a = 7.29 m / s²
now we can look for the tension
T = ½ M a
T = ½ 0.110 7.29
T = 0.40 N
A simple generator is used to generate a peak output voltage of 23.0 V . The square armature consists of windings that are 5.1 cm on a side and rotates in a field of 0.500 T at a rate of 55.0 rev/s .
How many loops of wire should be wound on the square armature?
Answer: 51
Explanation:
Given
Output is 23 V
The square armature side is [tex]a=5.1\ cm[/tex]
Magnetic field [tex]B=0.5\T[/tex]
Rate of revolution [tex]n=55\ rev/s[/tex]
Angular speed
[tex]\omega =2\pi n\\\omega=2\pi \times 55=110\pi\ rad/s[/tex]
Peak voltage is given by
[tex]E_{peak}=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=\dfrac{E_{peak}}{B\omega A}\\\\N=\dfrac{23}{0.5\times 110\pi\times (0.051)^2}\approx 51[/tex]
So, there are approximate 51 loops
What is the height of a copper cylinder ( ρCu = 8.96 gcm-3) of diameter 10 cm with a mass of 10 kg ?
Answer:
h=0.142m=14.2cm
Explanation:
ρ=8.96 g/cm3=8960 kg/m3
d=0.1m
ρ=m/V - - >ρ=m/[π*((d/2)^2) *h] - - >
8960=10/[π*((0.1)^2) *h] - - >
h=0.142m
Two solenoids of equal length are each made of 2000 turns of copper wire per meter. Solenoid I has a 5.00 cm radius; solenoid II a 10.0 cm radius. When equal currents are present in the two solenoids, the ratio of the magnitude of the magnetic field BIalong the axis of solenoid I to the magnitude of the magnetic field BIIalong the axis of solenoid II, BI/BII, is
Answer:
BI/BII = 1
Explanation:
The magnetic field due to a solenoid is given by the following formula:
[tex]B = \mu nI\\[/tex]
where,
B = Magnetic Field due to solenoid
μ = permeability of free space
n = No. of turns per unit length
I = current passing through the solenoid
Now for the first solenoid:
[tex]B_1 = \mu n_1I_1 \\[/tex]
For the second solenoid:
[tex]B_2 = \mu n_2I_2\\[/tex]
Dividing both equations:
[tex]\frac{B_1}{B_2} = \frac{\mu n_1I_1}{\mu n_2I_2}\\[/tex]
here, no. of turns and the current passing through each solenoid is same:
n₁ = n₂ and I₁ = I₂
Therefore,
[tex]\frac{B_1}{B_2} = \frac{\mu nI}{\mu nI}\\[/tex]
BI/BII = 1
An aircraft has to fly between two cities, one of which is 600.0 km north of the other. The pilot starts from the southern city and encounters a steady 100.0 km/h wind that blows from the northeast. The plane has a cruising speed of 236.0 km/h in still air. In what direction (relative to east) must the pilot head her plane
Answer:
72.57° North of east
Explanation:
From the given information:
We can compute the velocity plane that is related to the ground in air in the North direction as;
[tex]v^{\to} _{PG} = v \\ \\ v^{\to} _{PG,x} = 0 \\ \\ v^{\to} _{PG,y} = v[/tex]
However, the velocity of the wind-related to the ground from the NorthEast direction is;
[tex]v^{\to}_{wG}=100 \ km/h \\ \\ \text{from North East} \\ \\ v_{wG,x} = (-100 \ km/h ) cos 45 = -70.7 km/h \\ \\ v_{wG,y} = (-100 \ km/h ) sin 45 = -70.7 km/h[/tex]
Now,
Since the plane is moving with a 236 km/h speed in the Northeast direction;
Then;
[tex]v^{\to} _{pw} = 236 \ km/h \\ \\ v^{\to} _{pw.x} = (236 m/s) cos \theta \\ \\ v^{\to} _{pw,y} = (236\ m/s) sin \theta \\ \\ v_{pG,x} = v_{pw,x} + v_{w G,x} \\ \\ \implies 0 = (236 \ km/h) sin \theta -( 70.7 \ km/h) \\ \\ \implies cos \theta = \dfrac{70.7 \ km/h}{236 \ km/h} \\ \\ \theta = cos^{-1} (0.2996) \\ \\ \mathbf{\theta = 72.57}[/tex]
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result of a directed force combined with a friction force. As a result of the applied torque the angular speed of the wheel increases from 0 to 10.3 rad/s. After 6.10 s the directed force is removed, and the wheel comes to rest 60.6 s later.
(a) What is the wheel's moment of inertia (in kg m2)? kg m
(b) What is the magnitude of the torque caused by friction (in N m)? N m
(c) From the time the directed force is initially applied, how many revolutions does the wheel go through?
______ revolutions
Answer:
[tex]21.6\ \text{kg m}^2[/tex]
[tex]3.672\ \text{Nm}[/tex]
[tex]54.66\ \text{revolutions}[/tex]
Explanation:
[tex]\tau[/tex] = Torque = 36.5 Nm
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\omega_f[/tex] = Final angular velocity = 10.3 rad/s
t = Time = 6.1 s
I = Moment of inertia
From the kinematic equations of linear motion we have
[tex]\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2[/tex]
Torque is given by
[tex]\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2[/tex]
The wheel's moment of inertia is [tex]21.6\ \text{kg m}^2[/tex]
t = 60.6 s
[tex]\omega_i[/tex] = 10.3 rad/s
[tex]\omega_f[/tex] = 0
[tex]\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2[/tex]
Frictional torque is given by
[tex]\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}[/tex]
The magnitude of the torque caused by friction is [tex]3.672\ \text{Nm}[/tex]
Speeding up
[tex]\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}[/tex]
Slowing down
[tex]\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}[/tex]
Total number of revolutions
[tex]\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}[/tex]
[tex]\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}[/tex]
The total number of revolutions the wheel goes through is [tex]54.66\ \text{revolutions}[/tex].
Acellus
A motion sensor emits sound, and
detects an echo 0.0115 s after. A
short time later, it again emits a
sound, and hears an echo after
0.0183 s. How far has the
reflecting object moved?
Help Resources
(Speed of sound = 343 m/s)
(Unit = m)
Answer:
1.17m
Explanation:
The formula to find distance is d=vt/2
This problem is asking for how far the reflecting object has moved so you need to find the distance from the motion sensor at both times.
(343)(0.115) / (2) = 1.97
(343)(0.0183) /(2) =3.14
After that, all you have to do is find the difference so
3.14 - 1.97
= 1.17
PLEASE HELP and actually help plz
The position of masses 4kg, 6kg, 7kg, 10kg ,and 3kg are (0,1), (4,2), (3,5), (5,6), and (-2,4) respectively. Where must you place a mass of 13kg if you want the center of mass to be at (-1,-3)?
Answer:
iEvaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)
Explanation:
What is characteristic of a good insulator?
A. Electrons are usually not moving at all.
B. Electrons are free to move around.
C. Electrons are semi-free to move around.
D. Electrons are tightly bound to the nuclei.
Answer:
D. Electrons are tightly bound to the nuclei.
Explanation:
In an insulator, the electrons of the outer most shell are bound with a very high electrostatic forces coming from the nucleus of each atom so electrons cannot flow around all atoms making up the material as in a conductor.
The characteristic of a good insulator is Electrons are tightly bound to the nuclei. (option d)
In a good insulator, electrons are tightly bound to the nuclei of their atoms. This means that they are not free to move around within the material, unlike conductors where electrons are relatively loosely bound and can move freely. Due to this strong binding, electrons in insulating materials cannot carry an electric charge or energy easily from one atom to another.
When an electric field is applied to an insulator, the electrons may experience a small displacement within their respective atoms, but they generally do not move from one atom to another or flow through the material like they would in a conductor. As a result, insulators prevent the flow of electric current and are used to isolate or protect conductive elements from accidental contact.
So, the correct answer is D. Electrons are tightly bound to the nuclei.
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A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal disk is placed in contact with a wall, and the disk comes to rest after 10s. Which of the following situations associated with linear impulse is analogous to the angular impulse that is described?
a. A 3kg block is initially at rest. An applied force of 3N is applied to the block, but the block does not move.
b. A 3kg block is initially at rest. A net force of 3N is applied to the block until it has a speed of 10m/s.
c. A 3kg block is initially traveling at 10m/s. An applied force of 3N is applied to the block in the direction of its velocity vector for 10s.
d. A 3kg block is initially traveling at 10m/s. The block encounters a 3N frictional force until the block eventually stops.
Answer:
D
Explanation:
From the information given:
The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]
Disk radius (r) = 0.2 m
The block Initial velocity is:
[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]
Change in the block's angular speed is:
[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]
However, on the disk, moment of inertIa is:
[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]
The time t = 10s
∴
Frictional torques by the wall on the disk is:
[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]
Finally, the frictional force is calculated as:
[tex]F = \dfrac{T}r{}[/tex]
[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]
using the human species as an example, explain what is meant by variation of traits
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 3.00 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling ?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m
Explanation:
a) to find the velocity of the wave let us use the relation
v = λ f
the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength
λ = x
λ = x
the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period
T / 2 = t
T = 2t
period and frequency are related
f = 1 / T
f = 1 / 2t
we substitute
v = x / 2t
v = 5.50 / 2 3
v = 0.9167 m / s
b) the amplitude is the distance from a maximum to zero
2A = y
A = y / 2
A = 0.700 / 2
A = 0.350 m
c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same
v = 0.9167 m / s
d) the amplitude is
A = 0.500 / 2
A = 0.250 m
I WILL REPORT YOU IF YOU DON'T ANSWER QUESTION OR IF YOU PUT A LINK
Which of the following statements are true
What is the difference between a positively and negatively charged object?
Answer:
Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.
The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.
What are Charged objects?
Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.
A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.
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A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00 m long incline that is at 25.0 degrees to the horizontal, and it is then released to roll straight down. Assuming mechanical energy conservation, calculate the moment of inertia of the can if it takes 1.50 s to reach the bottom of the incline. Which pieces of data, if any, are unnecessary for calculating the solution
Answer:
I = 1.093 x 10⁻⁴ kg.m²
Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.
Explanation:
The can which is filled with the soup can be modelled as a solid cylinder. The moment of inertia of this solid cylinder about its axis of rotation can be given by the following formula:
[tex]I = \frac{1}{2}mr^2[/tex]
where,
I = moment of inertia of can = ?
m = mass of can with soup = 215 g = 0.215 kg
r = radius of can = diameter/2 = 6.38 cm/2 = 3.19 cm = 0.0319 m
Therefore,
[tex]I = \frac{1}{2}(0.215\ kg)(0.0319\ m)^2 \\[/tex]
I = 1.093 x 10⁻⁴ kg.m²
Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.