The actual mechanical advantage of any machine is its__________divided by its_______________

A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400

slava [35]

Answer:

28,699m

Explanation:

The force to make the box move should be <u>μs.N=μs.m.g=m.</u><u>|</u><u>a</u><u>|</u>

then,

|a|=μs.g

Being

μs coefficient of static friction,

N the force made by the truck on the box caused by the gravity force,

m the mass,

g the acceleration of gravity

and a the acceleration of the truck.

$x = v \times t + \frac{1}{2} \times a \times {t}^{2}$

as the truck is stopping, the acceleration is negative. then,

$x = v \times t - \frac{1}{2} \times |a| \times {t}^{2}$

$|a| = v \div t \\ t = v \div |a|$

$x = v \times (v \div |a| ) - \frac{1}{2} \times |a| \times {(v \div |a|)}^{2}$

$x = v \times (v \div μs.g ) - \frac{1}{2} \times |a| \times {(v \div μs.g)}^{2}$

$x = \frac{{v}^{2}}{μs \times g} - \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} } = 28.699m$

28,699m

7 0

3 years ago

SOUND IS A FORM OF WAVE.LIST AT LEAST 3 REASONS TO SUPPORT IDEA THAT SOUND IS A WAVE.

algol [13]

1 The question asks for a certain quantity of examples in a list (Name 6 factors that contributed to the start of World War I, What 3 subatomic particles constitute an atom? etc).
2 The question is academically precise and, therefore, indecisive in the wording (What are the 2 kinds of loading most professional engineers and academics in the field of engineering today generally consider to be relevant in most cases when considering typical types of structure usually made of common materials using well-understand methods?)
3 The question challenges the answerer to defend a position as opposed to merely rattling off a list based on knowledge alone, thereby invoking higher levels of Bloom's Taxonomy. (What are 4 arguments that could be used to defend arguments made by the physicists of the day that electromagnetic waves must move through an illusive substance called 'the ether?)

6 0

3 years ago

IM DESPERATE PLS HELP ME I DONT UNDERSTAND THIS! PLEASE PLEASE PLEASE ANSWER BACK

OlgaM077 [116]

Answer:

The speed with which the man flies forward is 5.5 m/s

Explanation:

The mass of the man = 100 kg

The mass of the scooter = 10 kg

The speed with which the man was traveling on the scooter = 5 m/s

The speed of the scooter after it hits the rock = 0 m/s

Let v represent the speed with which the man flies forward

The formula for momentum, P, is P = Mass × Velocity

The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;

The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s

The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v

When the momentum is conserved, we have;

550 kg·m/s = 100 kg × v

∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.

The speed with which the man flies forward = v = 5.5 m/s

8 0

3 years ago

An astronomical unit (A.U.) is 1 point A) a term for defining the luminosity of a star B) the average distance from the Earth to

Elenna [48]

Answer:

B) the average distance from the Earth to the Sun

Explanation:

7 0

2 years ago

A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the

never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2= 91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

$\rm m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2$

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0

2 years ago

The actual mechanical advantage of any machine is its__________divided by its________________

The actual mechanical advantage of any machine is itsdivided by its______