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Iteru [2.4K]
3 years ago
5

What is the value of Vo in the

Physics
1 answer:
notsponge [240]3 years ago
5 0

Answer: 60m/s

Explanation:

From the diagram:

Θ = 30°

Vertical resolution (y-axis) :

Voy = VoSinΘ

g in the upward direction = negative (-) = - g

Vfinal = 0

Distance (H) traveled along y =

Time taken to reach maximum height :

From v = u + at

0 = usinΘ - gt

gt = usinΘ

t = usinΘ / g

Horizontal resolution:

S = ut + 1/2at^2

Substituting t = usinΘ / g ; Voy = usinΘ

S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2

S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)

S = (u^2sin^2Θ) / 2g

Now if S = maximum height = 45m

Then,

45 = [Vo^2sin^2(30°)] / 2(10)

45 =[ Vo^2 * (0.5)^2] / 20

45 =( Vo^2 * 0.25) / 20

20 * 45 = Vo^2 * 0.25

900 / 0.25 = Vo^2

3600 = Vo^2

Vo = sqrt(3600)

Vo = 60m/s

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A student lifts 100 N of books 1 m to his shoulder. He walks forward
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A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

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Explanation:

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Mass of the cart, m_1 = 300\ g = 0.3\ kg

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Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

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4 0
3 years ago
The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov
Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

4 0
3 years ago
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