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3 years ago

Help with 1 2 and 3 please

Physics

1 answer:

geniusboy [140]3 years ago

6 0

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. $I = \frac{q}{t}$ Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3. So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

$\frac{p1L1}{A1} = \frac{p2L2}{A2}\\$

We are looking for L2 so we can isolate using algebra to get:

$\frac{A2(\frac{P1L1}{A1}) }{P2} = L2$

If you fill in those values you get 0.0205

or 2.05 cm

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Why are time machines not invented?

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Have you seen back to the future?

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3 years ago

If a 3.1 kg ring of granite decreases in temperature 17.9C. What is the

steposvetlana [31]

I think it’s C but I’m not sure:/

6 0

3 years ago

One point charge has a magnitude 5.4x10-7 C. A second charge 0.25 meters away has a magnitude of 1.1x10-17 C. What is the electr

choli [55]

The electric force on the charges will be equal and opposite in nature and the magnitude will be 8.5536 × 10⁻¹³ Newton.

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

In question we have given

value of q1 = 5.4 x 10⁻⁷ C

value of q2 = 1.1 x 10⁻¹⁷ C

distance between the (r) = 0.25 m

Applying the formula

F = ( K x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 10⁹ ) x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 5.4 × 1.1) × ( 10⁹ × 10⁻⁷ x 10⁻¹⁷) )/ ( 6.25 × 10⁻² )

F = ( 53.46 × 10⁻¹⁵) / ( 6.25 × 10⁻² )

F = 8.5536 × 10⁻¹³ Newton

Electrostatic force = 8.5536 × 10⁻¹³ Newton

So, The point charges are possessing equal and opposite electrostatic force of magnitude 8.5536 × 10⁻¹³ Newton.

Learn more about Electrostatic Force here:

brainly.com/question/23121845

#SPJ10

8 0

2 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago

A 397-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.

AleksAgata [21]

Answer:

h = 14.4 m

Explanation:

The height can be calculated by energy conservation:

$K_{r} + K_{t} - W = E_{p}$

<u>Where</u>:

W: is the work

$E_{p}$ : is the potential energy

$K_{r}$ : is the rotational kinetic energy

$K_{t}$ : is the transitional kinetic energy

Initially, the wheel has rotational kinetic energy and translational kinetic energy, and then when stops it has potential energy.

$K_{r} + K_{t} - W = E_{p}$

$\frac{1}{2}I\omega_{0}^{2} + \frac{1}{2}mv^{2} - W = mgh$

<u>Where</u>:

I: is the moment of inertia = 0.800 mr²

ω₀: is the angular speed = 25.0 rad/s

m: is the mass = P/g = 397 N/9.81 m*s⁻² = 40.5 kg

v: is the tangential speed = ω₀r²

Now, by solving the above equation for h we have:

$h = \frac{\frac{1}{2}(I\omega_{0}^{2} + mv^{2}) - W}{mg}$

$h = \frac{\frac{1}{2}(I\omega_{0}^{2} + m(\omega_{0}*r)^{2}) - W}{mg}$

$h = \frac{\frac{1}{2}(0.800*40.5 kg*(0.600 m)^{2}*(25.0 rad/s)^{2} + 40.5 kg*(25.0 rad/s*0.600 m)^{2}) - 2500 J}{40.5 kg*9.81 m/s^{2}} = 14.4 m$

Therefore, the height is 14.4 m.

I hope it helps you!

4 0

3 years ago