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xxTIMURxx [149]
3 years ago
10

Which is usually greater for a pair of surfaces, the coefficient of static friction or the coefficient of sliding friction? why

Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Explanation :

Static friction is the frictional force between two objects that are at rest. While sliding friction is the frictional force between two objects in contact and are sliding w.r.t each other.

Static friction is usually greater than sliding friction because in static friction the contact forces is more and the interlocking between objects is tight as compared to sliding friction.    

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How are the energy conversions of machines similar to the energy conversions in your body
sleet_krkn [62]

Answer:

Explained below

Explanation:

When we eat food, our body gets chemical energy from it. Now, this chemical energy from the food is changed into some different energy forms that is useful to it. They include:

-Chemical to mechanical energy to aid in movement of muscles

- chemical to thermal energy to aid in regulating the body temperature.

- chemical to electrical energy to aid the brain in thinking.

Thus is similar to how a machine converts energy because machines also generate energy after being powered and convert to other forms of energy. For example, an alarm clock converts electrical energy to sound energy, hair dryer converts electrical energy to thermal/heat energy.

6 0
3 years ago
Can someone answer these
LenKa [72]

Four

Sometimes I think the creators of problems out to drawn and quartered. 60 g does not mean 60 grams. It means 60 * the acceleration due to gravity.

So the question really reads. The acceleration delivered by the air bag is 60 times that of a normal gravitational. This acceleration is delivered to the person where his mass is putting up a whole lot of resistance because he and his 75 kg are moving forward with the impact of the car. The 36 msec. has nothing to do with the problem.

The Force of the Air Bag is mass * a

F_airbag = mass * acceleration = 75 kg * 60 * 9.81 mass * acceleration = 44145 newtons

The answer is 4.41 * 10^4

Answer C

Five

This problem is governed by one formula that you sort of have to get out of your hat -- a piece of magic if you will.

Fg - Bf = m * a

Fg = the Force of gravity

Bf = the braking force

The mass of the rocket is derived from its weight

The acceleration is derived from one of your big 4 equations.

m of the rocket = 75600 / 9.81 = 7706 kg

The acceleration =

vi = 1 km/s = 1000 m/s

vf = 0

t = 2 minute * 60 sec/ min = 120 seconds

a = (vf - vi)/t = (0 - 1000 m/s) / 120 sec

a = - 8.333 m/s^2 The minus sign makes perfect sense. Remember the rocket is slowing down

The net downward force = mass * acceleration = - 7706 kg * - 8.333 m/s^2

The net force = - 64217 N

So going back to the problem's equation we have

Gravitational force - Braking Force = Net Force

Gravitational Force = 75600

Net Force = - 64217

Bracking force = ?

75600 - Bracking force = - 64217  Subtract 75600 from both sides

- Bracking force = - 64217 - 75600

- Braking force = - 139817

Braking force = 139817 N = 1.398 * 10^5 N

Braking Force = 1.4 * 10^5

Answer: Last One.

Six

The first thing you should do is derive a general formula for this problem.

The force pulling both masses down is M*g where g is the acceleration due to gravity.

The formula for this problem is

Mg = (m + M) * a

Now you need to solve for a

a =  [M/(M + m) ] * g

Look what is happening. is a smaller or larger than g? This is a question you should really pay attention to. If it was larger, everyone would have this system in their basement because you'd get more energy output than you put in. Something for nothing is always appealing.

So what's the answer? (I get to ask it. No one posing the question ever should).

A

A is incorrect. M never goes away. The acceleration may get very tiny, but there always is some acceleration.

B must be true. It is just what I finished saying about A

C Who said anything about velocity? It's a red herring. If the velocity became 0 the acceleration would have to turn minus. This answer sounds good, but sounds good doesn't make it right. C is wrong.

D The acceleration does not remain constant no matter what. The answer to A still applies. So D is wrong.

4 0
3 years ago
Riparian zones are attractive to livestock because _____. they are far from human presence they contain lush vegetation that liv
podryga [215]

The climate is more temperate here.

7 0
3 years ago
Read 2 more answers
HW 5.2.A 5.00-kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley whenit collides with and sticks to an
Allisa [31]

The concept used to solve this problem is the conservation of momentum and the conservation of energy.

For conservation of the moment we have the definition:

m_1v_1 = (m_1+m_2)v_f

Where,

m = Mass

v_1 = Initial velocity for object 1

v_f = Final velocity

Replacing the values we have to,

m_1v_1 = (m_1+m_2)v_f

5*12=(5+5)v_f

v_f = 6m/s

By conservation of energy we know that the potential energy is equal to the kinetic energy then

mgh = \frac{1}{2} m(v_f^2-v_i^2)

gh = \frac{1}{2} v_f^2

h = \frac{1}{2} g*v_f^2

h = \frac{1}{2} (9.8)(6)^2

h = 1.837m

Therefore after the collision the height when the combined chinks will go is 1.837m

3 0
3 years ago
Starting from rest at a height equal to the radius of the circular track, a block of mass 24 kg slides down a quarter circular t
LekaFEV [45]

Answer:

The work done against friction is 372 joules

Explanation:

It is given that,

Mass of block, m = 24 kg

Radius of the track, r = 15 m

Acceleration due to gravity, a=9.8\ m/s^2

If the kinetic energy of the block at the  bottom of the track is, 3900 J

Let P is the work done against friction. It is given by :

P=mgh

Here, h = r

P=24\ kg\times 9.8\ m/s^2\times 15\ m

P = 3528 J

Since it ends up with 3900 J, the work done is given by or the lost in energy will be :

W = 3900 - 3528

W = 372 joules

So, the work done against friction is 372 joules. Hence, this is the required solution.

6 0
3 years ago
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