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Oksanka [162]
4 years ago
7

Listed in the Item Bank are key terms and expressions, each of which is associated with one of the columns. Some terms may displ

ay additional information when you click on them. Drag and drop each item into the correct column. Order does not matter.
ITEM BANK: Move to Bottom
A short, thick, cold wire is the best conductor.A shorter wire will allow electricity to move through at a higher rate than a longer wire.How well a material conducts current is an internal factor affecting resistance.If you double the length of a wire, you cut the resistance in half.If you double the thickness of a wire, you cut the resistance in half.Superconductors have no measurable resistance.The higher the temperature of the conductor, the lower the resistance.The resistance in a wire with less thickness is less.Thickness, length, and temperature are internal factors that affect resistance.When a light is first switched on, the light bulb's filament has a lower resistance than after it gives off light for awhile.
TRUE
FALSE
Physics
1 answer:
zavuch27 [327]4 years ago
7 0

Answer:

Explanation:

True                                                                              

How well a material conducts current is an internal factor affecting resistance.A shorter wire will allow electricity to move through at a higher rate than a longer wire.When a light is first switched on, the light bulb's filament has a lower resistance than after it gives off light for awhile.Superconductors have no measurable resistance.If you double the thickness of a wire, you cut the resistance in half.A short, thick, cold wire is the best conductor.

False

Thickness, length, and temperature are internal factors that affect resistance.The resistance in a wire with less thickness is less.The higher the temperature of the conductor, the lower the resistance.If you double the length of a wire, you cut the resistance in half.

You might be interested in
For a series circuit what is the terminal voltage of a battery or power supply equal to in terms of the potential difference or
trapecia [35]

Answer: V=IR

Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as

E = I(R + r) and by expanding, we have that E =IR + It

Where r is the internal resistance of the battery

I is the total current flowing in the circuit

R total load resistance in the circuit.

E is the total emf of the circuit.

The total emf is the sum of 2 separate voltages.

"IR" which is the terminal voltage and "Ir" which is the loss voltage.

The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.

7 0
3 years ago
The burner on a stove is 325° f. given that the burner emits electromagnetic radiation as a blackbody, what is the maximum wavel
lesantik [10]
To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin. 

C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C

C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K

λmax = 2898/435.78 = <em>6</em><em>.65 μm</em>

5 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,
sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

t=15.2 s is the time it takes to the child to reach the bottom of the slope

V_{o}=0 is the initial velocity (the child started from rest)

\theta=15\° is the angle of the slope

d is the length of the slope

Now, the Force exerted on the sled along the ramp is:

F=ma (1)

Where m is the mass of the sled and a its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

F=mg sin \theta (2)

Where g=9.8 m/s^{2} is the acceleration due gravity

Then:

ma=mg sin \theta (3)

Finding a:

a=g sin \theta (4)

a=9.8 m/s^{2} sin(15\°) (5)

a=2.5 m/s^{2} (6)

Now, we will use the following kinematic equations to find d:

V=V_{o}+at (7)

V^{2}=V_{o}^{2}+2ad (8)

Where V is the final velocity

Finding V from (7):

V=at=(2.5 m/s^{2})(15.2 s) (9)

V=38 m/s (10)

Substituting (10) in (8):

(38 m/s)^{2}=2(2.5 m/s^{2})d (11)

Finding d:

d=288.8 m

6 0
3 years ago
a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity
stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

V_{ox} = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = V_{ox}   t

80 = (v Cos60) (t)

t = 160/v                                         eq-1


consider the motion in vertical direction :

Y = vertical displacement = 20 m

V_{oy}  = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = V_{oy}  t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

4 0
3 years ago
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