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3 years ago

Do the data for the first part of the experiment support or

Physics

1 answer:

mrs_skeptik [129]3 years ago

7 0

Answer:

your the class suru and Sahiba and Sahiba are all guddu saifi and Mam aap Aman and Sahiba are not

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Work and energy both have the unit?

sveta [45]

The both have the unit (J) for Jules

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3 years ago

If you increase the resistance in a series circuit, ________________

GrogVix [38]

B) the current will decrease

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3 years ago

Find the pressure exerted on the floor by a 100N box whose bottom area is 40cm x 50cm.

Mazyrski [523]

Answer:

Pressure exerted = 500 Pa

Explanation:

The formula for pressure is as follows:

$Pressure = \frac{Force \space\ applied}{Area}$

In this case,

Force applied = 100N

Area = 40cm × 50cm = 2000cm² = 2000 × 10⁻⁴ = 0.2m²

Substituting these values into the formula:

Pressure = $\frac{100}{0.2}$

⇒ Pressure = 500 Pa

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2 years ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

3 years ago

Which three quantities can be used to calculate acceleration?

ikadub [295]

The correct option will be
D. Time, initial velocity and final velocity
The Formula can be written as,
Acceleration=Final velocity-Initial Velocity/Time

5 0

3 years ago