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3 years ago

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

Physics

1 answer:

Rasek [7]3 years ago

6 0

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

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3 years ago

10 kg mass sliding along a horizontal rough surface at speed 30 m/s and rests in 6 seconds. what is the coefficient of friction

Aleks [24]

The coefficient of friction is 0.5.

<h3>What is Coefficient of friction?</h3>

The ratio of friction force to normal force is known as the coefficient of friction (COF), which has no dimensions. Those materials are said to as lubricous if their COF is less than 0.1. Surface roughness and COF are dependent on the composition of the materials.

Initial velocity, u=30m/s

time taken = 6s

Final velocity, v=0

Acceleration of friction, a=μg=10μms⁻²

Apply first kinematic equation of motion,

v - u = at

0 - 30 = (-10μms⁻²) × 6

-30 = -60μ

μ = 1/2

μ = 0..5

Hence, coefficient of friction is 0.5

to learn more about Coefficient of friction go to - brainly.com/question/14121363

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Answer:

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Explanation:

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3 years ago

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.