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2 years ago

Which equation relates charge, time, and current?

Physics

1 answer:

san4es73 [151]2 years ago

8 0

Answer:

The first one

Explanation:

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Which item is heavier, 50 kg of cotton or 50 kg of iron?

Kipish [7]

Answer:

They weight the same

Explanation:

6 0

2 years ago

A horizontal force of 90.0 N is required to push a 75kg object along a horizontal surface at a constant speed. What is the magni

Varvara68 [4.7K]

If the object is moving at a constant speed, acceleration is 0. So:

$F = ma

F = m\times 0

F = 0$

The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so

Friction = 90 N

6 0

3 years ago

A steel rope is used to lift a 26 kg slab of concrete from the ground to a height of 20 m. Assume that the rope moves the concre

Ne4ueva [31]

Answer:

5200 Joules

Explanation:

Work Formula:

W = F . D

W = (26.10) . 20

W = 260 . 20

W = 5200 Joules

5 0

3 years ago

A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a

Kryger [21]

Answer:

2583.9 N/C

Explanation:

Parameters given:

Outer diameter = 14 cm

Outer radius, R = 7cm = 0.07m

Inner diameter = 7 cm

Inner radius, r = 3.5 cm = 0.035m

Charge of washer = 8 nC = 8 * 10^(-9)C

Distance from washer, z = 33 cm = 0.33m

The electric field due to a washer (hollow disk) is given as:

E = k * σ * 2π [ 1 - z/(√(z² + R²)]

Where σ = charge per unit area

σ = q/π(R² - r²)

σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)

σ = 2.077 * 10^(-6) C/m²

=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]

E = 117.467 * 10^3 * (1 - 0.978)

E = 117.467 * 10^3 * 0.022

E = 2583.9 N/C

6 0

3 years ago

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

$Q=cm\Delta T$ (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and $\Delta T$ the change in temperature of the sample. So, if we solve (1) for

Sample A:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}$

$\Delta T=-\frac{Q}{16744}$

Sample B:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}$

$\Delta T=-\frac{Q}{5400}$

Sample C:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}$

$\Delta T=-\frac{Q}{9450}$

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression $-\frac{Q}{cm}$ that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of $\Delta T$ implies the temperature is dropping)

8 0

2 years ago