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3 years ago

Hurryyyyy plzzzzzz..........the one in the middle...

Physics

1 answer:

zheka24 [161]3 years ago

5 0

Answer:

Friction force is independent of the direction of the contacting surfaces

Explanation:

It can go any way depending on how much force is being out on it.

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A uniform square crate is released from rest with corner D directly above A and it rotates about A until its corner B impacts th

Tju [1.3M]

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4 years ago

As light shines from air to another medium, i = 30.0o, the light bends toward the normal and refracts at 22.0o. What is the inde

iVinArrow [24]

The answer is n = 1.3347

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4 years ago

Read 2 more answers

An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other

Umnica [9.8K]

The answer is c. +2.0 µC

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹ N*m/C²

It is given:

F = 7.2 N

d = 0.1 m = 10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Q2 = ?

Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²

7.2 = 36 × 10³ * Q2 / 10⁻²

Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶

Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

5 0

3 years ago

A commercial jet liner takes off with an average acceleration of 3 g. How long does it take to reach the end of its runway which

ANEK [815]

Answer:

The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

Explanation:

Given;

average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²

distance traveled by the commercial Jet liner, s = 1542 m

The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocity of the commercial Jet liner = 0

s = 0 + ¹/₂at²

s = ¹/₂at²

2s = at²

$t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2 \times 1524}{29.4} } \\\\t = 10.18 \ s$

Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

6 0

3 years ago

What is the current if 4C of charge passes in 2 s?

julia-pushkina [17]

Answer:

I hope 2 amperes of current passes

8 0

3 years ago