3 years ago

A driver can exceed the postage maximum speed limit in a work zone T or F?

Physics

1 answer:

PIT_PIT [208]3 years ago

6 0

That statement is physically and grammatically true but legally false.

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A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time el

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B is the answer to this

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3 years ago

How is air resistance similar to gravity? give me two ways.

OLga [1]

Answer:

1. they both act on an object in free fall

Explanation:

2. both help determine how fast the object will accelerate

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2 years ago

Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If

cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

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½mv² = 375 × ¾

½mv² = 281.25 J

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2 years ago

Water, H2O ,and methane ,CH4, are both covalent substances . why is water a liquid at room temperature while methane is a gas ?

stiv31 [10]

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2 years ago

Read 2 more answers

If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t

Aleksandr-060686 [28]

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

$B = \frac{\mu I}{2 \pi r}$

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

$Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT$

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

7 0

2 years ago