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bearhunter [10]
3 years ago
12

Need help ASAPJenna made an electric circuit as seen in the picture. She placed a thermometer near her light bulb. After the lig

ht bulb was turned on for an hour, what did she MOST LIKELY observe?
A) The bulb glowed brighter over time.
B) The temperature in the thermometer went up.
C) The temperature in the thermometer went down.
D) Both the light and temperature remained the same over time.
Physics
2 answers:
zepelin [54]3 years ago
7 0
B) The temperature in the thermometer went up.
maks197457 [2]3 years ago
7 0
B) The temperature in the thermometer went up.
Hope it helps :D
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nexus9112 [7]

Absolute zero is not about numbers.  It's about temperature, and the
motion of molecules in gases. 

You know that the temperature we feel with our skin is the result of the
average speed of all the tiny molecules zipping around or vibrating in
the solid, liquid, or gas.

The faster they're all moving, the warmer the substance feels to us. 
The slower they're all moving, the cooler the substance feels to us.

When molecules slow down to zero and lose all of their kinetic energy,
that temperature is what we call 'absolute zero' ... if they're not moving
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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
irinina [24]

288.51 N is  the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The  horizontal component of force + the vertical component of force  

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51  N.

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7 0
1 year ago
A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex
irina1246 [14]

a)

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

F_{B_x}=0

F_{B_y}=0

b)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=0

F_{B_y}=3.21\cdot 10^{-3}N (+z axis)

c)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=3.21\cdot 10^{-3} N (+y axis)

F_{B_y}=3.21\cdot 10^{-3}N (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

F=qE

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

q=+4.9\mu C=+4.9\cdot 10^{-6}C is the charge

E_x=+242 N/C is the electric field, along the x-direction

So the electric force (along the x-direction) is:

F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

B is the magnetic field

\theta is the angle between the directions of v and B

Here the charge is stationary: this means v=0, therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

F_{B_y}=qvB_y

where:

q=+4.9\cdot 10^{-6}C is the charge

v=345 m/s is the velocity

B_y = +1.9 T is the magnetic field

Substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

For the field B_x, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

F_{B_x}=qvB_x

And by substituting,

F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field B_y, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

F_{B_y}=qvB_y

And by substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0
3 years ago
Which of the following is an ecosystem
iogann1982 [59]

A it is Green plants.

or they all

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