Without friction, what happens? Check ALL

If a person wants to move a heavy box across the room, what is required to get the box in motion? apply any force apply a force

garik1379 [7]

If a person want to move a <span>heavy box across the room, he must apply a force that is greater than starting frictional force acted on the heavy box in order to get the box in motion. the frictional force is equal to coefficient of friction times the normal force, and normal force is approximately the weight of the object. so the force that must be applied must be greater than the wieght of the object.</span>

4 0

3 years ago

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What do we mean by primitive material? How can we tell if a meteorite is primitive?

zimovet [89]

Answer:

Primitive material was formed in the early stage of the solar system, before planets cooled off enough to differentiate elements of different density. It was not subject to great heat or pressure after it formed.

We can identify primitive meteorites by their composition, primitive meteorites are usually undifferentiated stones, with some metallic grains mixed in. Some primitive meteorites are darker, carbonaceous stones.

4 0

3 years ago

Why is a living thing a system?

Wittaler [7]

Any living thing has multiple things going on in it's body at one time and all are working on separate or similar tasks in the organism, therefore they work as a system.

4 0

3 years ago

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A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of

lutik1710 [3]

Answer:

6900 m/s

Explanation:

The mass of the rocket is:

m = 330000 − 280000 (t / 250)

m = 330000 − 1120 t

Force is mass times acceleration:

F = ma

a = F / m

a = F / (330000 − 1120 t)

Acceleration is the derivative of velocity:

dv/dt = F / (330000 − 1120 t)

dv = F dt / (330000 − 1120 t)

Multiply both sides by -1120:

-1120 dv = -1120 F dt / (330000 − 1120 t)

Integrate both sides. Assuming the rocket starts at rest:

-1120 (v − 0) = F [ ln(330000 − 1120 t) − ln(330000 − 0) ]

-1120 v = F [ ln(330000 − 1120 t) − ln(330000) ]

1120 v = F [ ln(330000) − ln(330000 − 1120 t) ]

1120 v = F ln(330000 / (330000 − 1120 t))

v = (F / 1120) ln(330000 / (330000 − 1120 t))

Given t = 250 s and F = 4.1×10⁶ N:

v = (4.1×10⁶ / 1120) ln(330000 / (330000 − 1120×250))

v = 6900 m/s

4 0

3 years ago

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a

Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

$w_1g = 1.4*9.81*y*sin30$ = - 6.867y

Workdone by gravity on the mass of the suspended block is:

$w_2g = 0.06*9.81$ = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y² - 6.867y + 0.5886 = 0.0588

-20 y² - 6.867y + 0.5886 - 0.0588

-20 y² - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y² + 6.867y - 0.5298 = 0

Using the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where;

a = 20 ; b = 6.867 c= - 0.5298

$\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$

$= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$ = 0.0649 OR −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0

3 years ago