All categories

3 years ago

2 Points

Physics

2 answers:

liberstina [14]3 years ago

8 0

A- electrical energy to kinetic energy

max2010maxim [7]3 years ago

8 0

Answer:

electric to kinetic like the other dude said

Explanation:

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A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t

Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

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3 years ago

What is an non example of Cultural diffusion

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Beliefs is a non example

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3 years ago

A 10 N force and a 15 N force are acting from a single point in opposite directions. What additional force must be added to prod

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Answer:

5 N acting in the same direction as the 10 N force

Explanation:

10+5=15

15=15

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3 years ago

A force of 185 n is needed to keep a small boat moving at 2.39 m/s. what is the power required to keep the boat moving at the st

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We know, P = F . ΔV

P = 185 * 2.39 = 442.15 watt

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3 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago