Can light cause the plastic to melt? Why or why not? Does it matter what type of light he shines on the plastic

A certain laboratory experiment requires an aluminum wire of length of 30.0 m and a resistance of 3.80 ω at 20.0°c. what diamete

Tom [10]

Answer:
diameter = 5.316 * 10^-4 meters

Explanation:
The resistance of the wire can be calculated using the following rule:
R = ΡL / A
where:
R is the resistance = 3.8 ohm
Ρ is the resistivity = 2.82 * 10^−8 ω · m
L is the length of the wire = 30 m
A is the area that we want to calculate
Substitute with the givens in the above equation to get the area as follows:
3.8 = (2.82 * 10^−8 * 30) / A
A = 2.22 * 10^-7 meters^2
Now, the area of the circle is calculated as follows:
Area = pi * (radius)^2
2.22 * 10^-7 = pi * (radius)^2
(radius)^2 = 7.066 * 10^-8
radius = 2.658 * 10^-4 meters
Diameter is double the radius. This means that:
diameter = 2 * 2.658 * 10^-4 = 5.316 * 10^-4 meters

Hope this helps :)

5 0

3 years ago

You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different

Aliun [14]

Answer:

a) The equation is $(m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

b) Your velocity after collision is 2.64 m/s

c) The force you felt is 7392 N

d) you and your brother undergo an equal amount of acceleration

Explanation:

Your mass $m_{y}$ = 60 kg

your brother's mass $m_{b}$ = 30 kg

mass of the car $m_{c}$ = 80 kg

your initial speed $u_{y}$ = 0 m/s (since you've not started moving yet)

your brother's initial velocity $u_{b}$ = 3 m/s

your final speed $v_{y}$ after collision = ?

your brother's final speed $v_{b}$ after collision = ?

a) equations you need to use to figure out how fast you and your brother are moving after the collision is

$(m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

but $u_{y}$ = 0 m/s

the equation reduces to

$(m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

b) if your little brother reverses with velocity of 0.36 m/s it means

$v_{b}$ = -0.36 m/s (the reverse means it travels in the opposite direction)

then, imputing values into the equation, we'll have

$(m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

(30 + 80)3 = (60 + 80) $v_{y}$ + (30 + 80)(-0.36)

330 = 140 $v_{y}$ - 39.6

369.6 = 140 $v_{y}$

$v_{y}$ = 369.6/140 = 2.64 m/s

This means you will also reverse with a velocity of 2.64 m/s

c) your initial momentum = 0 since you started from rest

your final momentum = (total mass) x (final velocity)

==> (60 + 80) x 2.64 = 369.6 kg-m/s

If the collision lasted for 0.05 s,

then force exerted on you = (change in momentum) ÷ (time collision lasted)

force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N

d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s

your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2

your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s

his deceleration is (3 - 0.36)/0.05 = 52.8 m/s

you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost

7 0

3 years ago

Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber

puteri [66]

Answer:

C. A rubber rod and a glass rod charged this way have opposite charges on them.

Explanation:

When a rubber rod is rubbed against cat fur, it acquires a negative charge, it becomes negatively charged.

When you then try to bring two rubber rod's together, they repel because like charges repel.

Meanwhile, when you rub a glass rod against silk, it loses electrons to the silk material and becomes positively charged.

When you bring two positively charged glass rod's together, they repel, because like charges repel.

However, when you bring the rubber rod and a glass rod together, the attract each other because unlike/opposite charges attract.

5 0

3 years ago

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0

3 years ago

Choose all of the items that are incorrectly matched. A) inactivation gate — closed at restB) activation gate — open at restC) i

Feliz [49]

D) activation gate - opens during depolarization

4 0

4 years ago