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sleet_krkn [62]
3 years ago
10

How many Gigameters in 1600 meters

Physics
2 answers:
larisa [96]3 years ago
8 0
If you mean kilometers then it's 1.6

USPshnik [31]3 years ago
6 0
1.6e-6 gl on the rest of your physics its a killer for me too
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A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?
ipn [44]
Your answer would be A. You divide 96 by 16 to find the answer
3 0
4 years ago
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Kinetic energy is the energy of
Alina [70]

Kinetic energy is the energy of motion.

The formula for kinetic energy is given as

KE = (0.5) m v²

where m = mass of object , v = speed of object.

an object having some speed and mass will have kinetic energy while an object at rest will not have any kinetic energy since the speed of object at rest is zero.

Since at rest speed is zero. an object has kinetic energy only when it is in motion.

3 0
3 years ago
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The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve
solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d =  0.107 mm =  0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 /  4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0
3 years ago
A parallel-plate capacitor is formed from two 1.6 cm -diameter electrodes spaced 2.8 mm apart. The electric field strength insid
lidiya [134]

Answer:

12 nC

Explanation:

Capacity of the parallel plate capacitor

C = ε₀ A/d

ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate

Area of plate = π r²

= 3.14 x (0.8x 10⁻²)²

= 2 x 10⁻⁴

C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³

= 7.08 x 10⁻¹³

Potential difference between plate = field strength x distance between plate

= 6 x 10⁶ x 2.8 x 10⁻³

= 16.8 x 10³ V

Charge on plate = CV

=7.08 x 10⁻¹³ X 16.8 X 10³

11.9 X 10⁻⁹ C

12 nC .

8 0
3 years ago
Need help with this question. Thirty points.
Anon25 [30]

1)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2 \\\\\implies m = \dfrac{2E_k}{v^2} = \dfrac{2\times 4500}{30^2}=10~ kg\\ \\2)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12mv^2 \\\\\implies v^2 = \dfrac{2E_k}m\\\\\implies v  = \sqrt{\dfrac{2E_k}{m}}=\sqrt{\dfrac{2 \times 320}{\dfrac{20}{9.8}}}= 17.709 ~ ms^{-1}\\\\3)\\\\\text{Kinetic energy,}\\\\E_k=\dfrac 12 mv^2=\dfrac 12 \times 50 \times 10^2=2500 ~J\\\\\\

4)\\\\\text{Potential energy,}\\\\E_p =mgh = 5\times 9.8\times 1.5 = 73.5 ~J

5 0
2 years ago
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