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Masja [62]
3 years ago
15

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

the incline? show all work please​
Physics
1 answer:
Leona [35]3 years ago
6 0

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

wsin\theta-\mu F_n=ma That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

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hram777 [196]

Gravitational force is determined by mass

Answer: Option B

<u>Explanation:</u>

According to Universal law of gravity, the gravitational force is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between the objects.

            F=G \times \frac{m_{1} \times m_{2}}{r^{2}}

Where,

G – gravitational constant = \text { 6. } 67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}

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8 0
3 years ago
You are at the top of the Empire State Building on the 102nd floor, which is located 373 m above the ground, when your favorite
Liula [17]

Answer:

1.5 m

Explanation:

H = actual height of the superhero = ?

H₀ = height of the superhero as observed = 1.73 m

v = speed of the superhero = 0.50 c

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H = H_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}

Inserting the values

H = 1.73 \sqrt{1 - \left ( \frac{0.50 c}{c} \right )^{2}}

H = 1.5 m

3 0
3 years ago
The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t
maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

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Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s

Work done is equal to the change in kinetic energy. It can be given as follows :

W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J

So, the required work done is 32.99 J.

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The purpose of the defense is to <span>prevent the opposing offense from advancing the ball.

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