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Effectus [21]
3 years ago
7

You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler

has −14 µC of charge. The tissue has 5 g of mass. What is the minimum charge required to pick up the tissue paper?
Physics
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge,  equate electrostatic force to weight  

we have F = W

so\frac{KQq}{r^2} =mg

putting all value in equation,

=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8

solving for q

q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}

or q=1.4*10^{-9}C

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Answer:

C. Bacteria

Explanation:

Bacteria and archaea are the two domains which consist of organisms with prokaryotic cells. Prokaryotic cells are cells without nuclei or membrane-bound organelles.

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How does the law of conservation of energy relate to the system shown in this model
egoroff_w [7]

Answer: The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another. Another way of stating this law of chemistry is to say the total energy of an isolated system.

Explanation:

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The weightlifter's internal store of energy decreased when he lifted the bar.
Mumz [18]

Answer:

The energy returns to the weightlifter's muscles, where it is dissipated as heat.

Explanation:

The energy returns to the weightlifter's muscles, where it is dissipated as heat. As long as the weightlifter controls the weight's descent, their muscles are acting as an overdamped shock absorber, as if the weight were sitting on a piston containing very thick fluid, slowly compressing it downward (and slightly heating up the fluid in the process). Since muscles are complicated biological systems and not simple pistons, they require metabolic energy to maintain tension throughout the controlled descent, so the weightlifter feels like they're putting energy into the weight, even though the weight's gravitational potential energy is being converted into heat within the lifter's muscles.

5 0
3 years ago
Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How
grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (\Delta s), in meters, can be determined by the following expression:

\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2} (1)

Where:

v_{o} - Initial velocity, in meters per second.

t - Time, in seconds.

a - Acceleration, in meters per square second.

If we know that v_{o} = 5\,\frac{m}{s}, t = 6\,s and a = 2\,\frac{m}{s^{2}}, then the box displacement after 6 seconds is:

\Delta s = 66\,m

The box displacement after 6 seconds is 66 meters.

5 0
3 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

4 0
3 years ago
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