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Delvig [45]
2 years ago
10

A 2kg block slides down a ramp which is at an incline of 25 degrees. Friction force is 4.86N. At what incline will the box slide

at a constant velocity
Physics
1 answer:
ehidna [41]2 years ago
6 0

The coefficient of kinetic friction between the block and the incline will be 0.133.

<h3>What is the cofficient of kinetic friction?</h3>

The ratio of the greatest kinetic friction force (F) between the surfaces in contact during movement begins to the normal (N) force is the coefficient of kinetic friction.

The complete question is;

"A block slides down an incline of an angle of 26° with an acceleration of 2.5 m/s^2. What is the coefficient of kinetic friction between the block and the incline?"

The given data in the problem is;

The coefficient of kinetic friction,(μ)=?

The minimum angle is such that the ladder may not slip is 25 degrees.

The coefficient of kinetic friction between the block and the incline is;

tanΘ=μ

tan 25=μ

tan 25°=-0.133

μ=-0.133

Hence the coefficient of kinetic friction between the block and the incline will be 0.133.

To learn more about the cofficient of static friction refer to;

brainly.com/question/17237604

#SPJ1

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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1
My name is Ann [436]

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons}  = 4.41294 Mev/nucleon

So approximately 4.41294 Mev/necleon


3 0
3 years ago
The picture shows a loop of wire rotating in a magnetic field in a generator. Determine the direction of the current through the
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I believe the answer is b
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3 years ago
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If the white car experiences a change in momentum equal to 105,000 kg*m/s and is in contact with the red car for 5 seconds befor
larisa86 [58]

Answer:

21000 N

Explanation:

From the question given above, the following data were obtained:

Change in momentum = 105000 kg.m/s

Time = 5s

Force =?

Force is related to momentum and time according to the following formula:

Force = Change in momentum / time

With the above formula, we can calculate the force the white car experience during the collision. This can be obtained as illustrated below:

Change in momentum = 105000 kg.m/s

Time = 5s

Force =?

Force = Change in momentum / time

Force = 105000 / 5

Force = 21000 N

Thus, the white car experience a force of 21000 N during the collision.

4 0
3 years ago
What is the mass of a large ship that has a momentum of 1.60×109kg·m/s, when the ship is moving at a speed of 48.0 km/h? (b) Com
erastova [34]

a) The mass of the ship is 1.2\cdot 10^8 kg

b) The ship has a larger momentum than the shell

Explanation:

a)

The momentum of an object is given by:

p=mv

where

m is the mass of the object

v is its velocity

For the ship in this problem, we have

p=1.60\cdot 10^9 kg m/s is the momentum

v=48.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=13.3 m/s is the velocity

Solving for m, we find the mass of the ship:

m=\frac{p}{v}=\frac{1.60\cdot 10^9}{13.3}=1.2\cdot 10^8 kg

b)

The momentum of the artillery shell is given by

p=mv

where

m is its mass

v is its velocity

For the shell in this problem,

m = 1100 kg

v = 1200 m/s

Substituting,

p=(1100)(1200)=1.32\cdot 10^6 kg m/s

So, we see that the ship has a larger momentum.

Learn more about momentum:

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6 0
3 years ago
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