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otez555 [7]
3 years ago
14

A=? m =3.4 kg F = 7kN a=f/m

Physics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

a=4,

m=7

F=

a=7so??sl

Explanation:

xddddddddddddd

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A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w
katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85  = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x  5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788  x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

4 0
3 years ago
Resonance in air column is used in which of these musical instruments
zvonat [6]

Answer: The correct answer is "trumpet".

Explanation:

Resonance occurs when the applied frequency on the object is equal to the natural frequency of the object.

Wind instruments use resonance in the air column to amplify the sound. It makes the sound louder. Sound is produced by using lips or vibrating reeds or buzzing sound into the mouthpiece.

Therefore, resonance in air column is used in trumpet musical instruments.

7 0
3 years ago
Read 2 more answers
A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect
navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is  one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.  

8 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
Which is an example of transfroming potential energy to kinetic​
Rudik [331]

Answer:

A ball being dropped to the ground

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