1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
allsm [11]
3 years ago
14

Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magnetic field. Explain whe

ther or not a current will be induced
Physics
2 answers:
Sonja [21]3 years ago
6 0

Answer:

A current will not be induced.  

Explanation:  

The wire is not part of a closed circuit.

The electrical charges have no place to flow to.

I am Lyosha [343]3 years ago
3 0

Answer: Sample Response:  A current will not be induced because the wire is not part of a closed circuit. The electrons have no place to flow to.

Explanation:

You might be interested in
During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient
ElenaW [278]

a) 1.58\cdot 10^6 J

b) 1.15\cdot 10^6 J

c) 0.43\cdot 10^6 J

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

U=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

g=9.8 m/s^2

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

U=(670)(9.8)(240)=1.58\cdot 10^6 J

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

W=F_f d

where

F_f is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

F_f=-\mu mg cos \theta

where

\mu=0.25 is the coefficient of friction

m = 670 kg is the mass of the rock

\theta is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}

Therefore, the work done by friction is:

W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J

So, the energy transferred to thermal energy is 1.15\cdot 10^6 J.

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

K_f = U_i -E_t

where here we have:

U_i=1.58\cdot 10^6 J is the potential energy of the rock at the top of the hill

E_t=1.15\cdot 10^6 J is the energy converted into thermal energy

Substituting, we find

K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

K_f=\frac{1}{2}mv^2

where

m is the mass of the rock

v is its final speed

In this problem, we have:

K_f=0.43\cdot 10^6 J is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s

7 0
3 years ago
Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on
rosijanka [135]

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}

s₁ = 1.1 m

Hence, the correct option is:

<u>b. 1.1 m</u>

4 0
2 years ago
The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

5 0
2 years ago
A 4.44 l container holds 15.4 g of oxygen at 22.55°c. what is the pressure?
padilas [110]
We will use the ideal gas equation:
PV = nRT, where n is moles and equal to mass / Mr
P = mRT/MrV
P = 15.4 x 8.314 x (22.55 + 273) / 32 x 4.44
P = 266.3 kPa
5 0
3 years ago
If you placed a dot on a world map to represent the location of every earthquake in the last 100 years, where would you see eart
Vedmedyk [2.9K]
If we are to place dots to teh places that have been struck by an earthquake these past 100 years, the dots would be concentrated in the east and southeast Asia region. This is because of the presence of the Pacific ring of fire. This is a major area in the Pacific Ocean where most of the earthquakes are likely to occur. 
6 0
2 years ago
Other questions:
  • Help with the following five problem please! For number 12 the teacher said 10 drops= 1 mm
    12·1 answer
  • Describe the motion of the object in Graph B of figure 11-2.
    15·2 answers
  • Which one of the following statements is true?
    9·1 answer
  • Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plate
    7·1 answer
  • Lena is studying the properties of light in a laboratory. If she increases the amplitude of the light waves she is studying, wha
    11·1 answer
  • HELP ME PLEASE
    15·1 answer
  • A box is placed on the floor.The area of the box in contact with the floor is 2.4m²Pressure exerted on the floor 16 newtons/m²Wo
    8·1 answer
  • I think this is physics but I just need someone to help me match the vocab plz
    5·1 answer
  • This downward force pulls indoor skydivers towards the Earth. *
    5·1 answer
  • A cannonball explodes in mid-air, fragmenting into several pieces. How does the total
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!