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Mila [183]
3 years ago
12

What initial speed v is required if the blocks m1 =2.5 kg and m2=1.5 kg are to travel a distance d =7.0cm before coming to rest?

Assume the coefficient of kinetic friction between m1 and the tabletop is ųk=0.21

Physics
1 answer:
ryzh [129]3 years ago
3 0

Answer:

<em>OPTRIMUM</em><em> </em><em>PRIDE</em><em> </em><em>URGH</em><em> </em><em>URGH</em><em> </em><em>URGH</em><em> </em>

Explanation:

AHHAAHAHAHAHA

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Why do you see objects when you shine a flashlight in a dark room?​
Afina-wow [57]

Answer:

We see objects in a dark room due to the emission of light photons which are sensitive to our eyes. Darkness is simply a terminology used to describe the absence of light. Visible light to human is a component of the electromagnetic spectrum. Our eyes have receptors that picks the photons which light releases

Explanation:

3 0
3 years ago
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Which of these forces determines the chemical properties of an atom? friction force nuclear force gravitational force electrical
zysi [14]
The answers is an electrical force.
Under normal conditions, atoms interact with each other via electrons that are furthest away from the nucleus. These electrons from the what is called the outer shell of the atom, electrons from the outer shell that can participate in chemical reactions are called valence electrons.

3 0
4 years ago
Using examples, explain why the first and second Newton laws of motion are significant for living organisms.
Triss [41]

Answer:

1) Newton's first law of motion states an object will remain at rest or in uniform will be in uniform motion in a straight line unless a force acts on it

2) Newton's second law states the acceleration of an object is directly proportional to the applied force acting on an object and inversely proportional to the mass of the object

Explanation:

1) With Newton's first law, we are able arrange things within a space and schedule meetings in time knowing that they will remain in place unless an external force changes their positions

2) An example of Newton's second law of motion is that small objects such as a ball are easily accelerated and can be given appreciable acceleration for flight by single, one time contact (such as kicking the ball) while larger objects such as a rock require sustained force application to change their location.

6 0
3 years ago
Circular Motion A 650-kg car moving at 8.5 m/s takes a turn around a circle with a radius of 48.0 m. Determine the acceleration
ollegr [7]

Answer:

Explanation:

The mass of the car doesn't matter because On a flat curve the mass of the car does not affect the speed at which it can stay on the curve. You would need the mass if you were solving the the centripetal force acting on the car, but not the acceleration.

a=\frac{v^2}{r} and filling in

a=\frac{(8.5)^2}{48.0} and we need 2 significant digits in our answer. That means that

a = 1.5 m/sec²

8 0
3 years ago
A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l
mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be \who_w = 1000 kg/m^3

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

F_s + F_b = W

Where F_s = kxN is the spring force, F_b = W_w = m_wg = \rho_w V_s g is the buoyancy force, which equals to the weight W_w of the water displaced by the submerged portion of the cylinder, which is the product of water density \rho_w, submerged volume V_s and gravitational constant g. W = mg is the weight of the metal cylinder.

kx + \rho_w V_s g = mg

The submerged volume would be the product of cross-section area and the submerged length x

V_s = Ax = \pi(d/2)^2x

Plug that into our force equation and we have

kx + \rho_w \pi(d/2)^2x g = mg

x(k + \rho_w g \pi d^2/4) = mg

x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m

6 0
3 years ago
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