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3 years ago

Newton first law of motion

Physics

2 answers:

Kaylis [27]3 years ago

6 0

Answer:

His first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... The amount of the change in velocity is determined by Newton's second law of motion.

tiny-mole [99]3 years ago

3 0

Answer:

It state that every body will remain at same condition (rest or motion) till external force acts on it ♥️

You might be interested in

A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m

Mariana [72]

Answer:

(a) 13.795 m/s.

(b) -3140.28 N.

Explanation:

(a) Using newton's equation of motion,

v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

Hence the velocity of the driver when he hits the water = 13.795 m/s.

(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

Substitute into equation 3

0² = 13.795²+2(2a)

0 = 190.30203 + 4a

-4a = 190.30203

a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

Substitute into equation 3

F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

7 0

3 years ago

How old is the universe

Virty [35]

Roughly 13.8 billion years old according to science

5 0

3 years ago

Read 2 more answers

A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the

shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

$Force, F = ma$

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

$a = 0.47 * 9.8 m/s^2$

$a = 4.606m/s^2$

We know,

$v^2 - u^2 = 2as$

where, v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

$u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s$

The car travels at a speed of 25m/s.

6 0

3 years ago

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$