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2 years ago

Imma straight male btw also wut are newtons law write a paragraph desricbing them

1 answer:

8 0

Third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

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. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

2 years ago

Mike has a mass of 97 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when

Ne4ueva [31]

Answer:

a) fr = 224.3 N , b) fr = 224.3 N , c) v = 198.0 m/s

Explanation:

a) For this exercise let's start by calculating the acceleration in the fall

v² = v₀² - 2 a (y-y₀)

When it jumps the initial vertical speed is zero

a = -v² / 2 (y-y₀)

a = -68 2/2 (1000-2000)

a = 2,312 m / s²

Let's use the second net law to enter the average friction force

fr = m a

fr = 97 2,312

fr = 224.3 N

b) let's look for acceleration

v² = v₀² - 2 a y

a = (v² –v₀²) / 2 (y-y₀)

a = (4² - 68²) / 2 (0-1000)

a = 2,304 m / s²

fr = m a

fr = 97 2,304

fr = 223.5 N

c) the speed of the wallet is searched with kinematics

v² = v₀² - 2 g (y-y₀)

v = √ (0-2 9.8 (0-2000))

v = 198.0 m/s

4 0

3 years ago

In a ________________ wave, such as a sound wave, the particles in the medium vibrate in the same direction that the wave travel

Varvara68 [4.7K]

The answer would be A, transverse.

6 0

2 years ago

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

2 years ago

Please help. I don’t understand this

skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

v (ave) = ∆x / ∆t

and under constant acceleration,

v (ave) = (v (final) + v (initial)) / 2

According to the plot, with ∆t = 4.0 s, we have v (initial) = 0 and v (final) = 10.0 m/s, so

∆x / (4.0 s) = (10.0 m/s) / 2

∆x = ((4.0 s) • (10.0 m/s)) / 2

∆x = 20.00 m

5 0

2 years ago