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Blizzard [7]
3 years ago
13

You push downward on a trunk at an angle 25° below the horizontal with a force of If the trunk is on a flat surface and the coef

ficient of static friction between the surface and the trunk is 0.61, what is the most massive trunk you will be able to move?

Physics
1 answer:
Sophie [7]3 years ago
7 0

Complete question is;

You push downward on a trunk at an angle 25° below the horizontal with a force of 750N. if the trunk is on a flat surface and the coefficient of static friction between the surface and the trunk is 0.61, what is the most massive trunk you will be able to move?

Answer:

The most massive trunk is about 81.3 kg

Explanation:

I've attached a free body diagram that depicts this question.

Where;

N = normal force on the trunk

m = mass of the trunk

W = weight of the trunk = mg

F = static frictional force

Using equilibrium of force in vertical direction, we obtain;

N = W + 750 Sin25

N = mg + 750 Sin25    - - - - (eq 1)

Now, we are given that Coefficient of static friction: μ = 0.61

static frictional force is given by the formula;

F = μN

Since N = mg + 750 Sin25, we now have;

F = (0.61) (mg + 750 Sin25)   - - - (eq 2)

Along the horizontal direction, for the trunk to move, force equation must be;

F = 750 Cos25

Thus, we now have;

750 Cos25 = 0.61(mg + 750 Sin25)

g = 9.81.

So,we now have ;

750 Cos25 = 0.61(m(9.81) + 750Sin25)

750 × 0.9063 = 0.61(9.81m + (750 × 0.4226))

Divide both sides by 0.61;

(750 × 0.9063)/0.61 = 9.81m + 316.95

1114.3 = 9.81m + 316.95

1114.3 - 316.95 = 9.81m

797.35 = 9.81m

m = 797.35/9.81

m = 81.3 kg

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3 0
3 years ago
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1. A 17.45-N force is applied to a 3.10-kg object to accelerate it rightwards. The object encounters 15.25 N of
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Answer:

please find attached file

Explanation:

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3 years ago
Consider the reaction.
BaLLatris [955]

Answer:

Explanation:

Consider the reaction.A(g)↽−−⇀Z(g)A

A

(

g

)

↽

−

−

⇀

Z

(

g

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The equilibrium constant,

K

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(g)

The energy of A-A bond in A is 98 kJ/mol. The energy of Z-Z bond in Z is 165 kJ/mol. Which statement about the reaction is correct?

1) The activation energy for the forward reaction, Reaction ↽−+X(g)AX(g), is smaller than the activation energy for the reverse reaction, Reaction ⇀+AX(g)↽−.

2) The amount of energy released when A and Z combine to form AX will be greater than the total energy needed to break both A-A and Z-Z bonds (above).

3) The amount of energy released when A and Z combine to form AX will be less than the total energy needed to break both A-A and Z-Z bonds.

4) The activation energy for the forward reaction, Reaction ↽−+X(g)AX(g), is greater than the activation energy for the reverse reaction, Reaction ⇀+AX(g)↽−.

5) The energy of A-A bond in A is greater than the energy of Z-Z bond in Z.

1) The activation energy for the forward reaction, Reaction ↽−+X(g)AX(g), is smaller than the activation energy for the reverse reaction, Reaction ⇀+AX(g)↽−.

It is correct that the activation energy for Reaction ↽−+X(g)AX(g) is smaller than that for Reaction ⇀+AX(g)↽−, because AX is less stable than A or Z, so it takes less energy to break the weaker bond in AX.

2) The amount of energy released when A and Z combine to form AX will be greater than the total energy needed to break both A-A and Z-Z bonds (above).

3) The amount of energy released when A and Z combine to form AX will be less than the total energy needed to break both A-A and Z-Z bonds.

Option (2) is incorrect because, though the formation of AX requires energy, which is not needed for the formation of A or Z; however, the total energy required to break both A-A and Z-Z bonds (above) is more than the amount released when A and Z combine to form AX.

Option (3) is incorrect because the amount of energy released when A and Z combine to form AX will be greater than the total energy needed to break both A-A and Z-Z bonds (above).

4) The activation energy for the forward reaction, Reaction ↽−+X(g)AX(g), is greater than the activation energy for the reverse reaction, Reaction ⇀+AX(g)↽−.

8 0
3 years ago
A light beam strikes a piece of glass at a 65.00 ∘ incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for
Leokris [45]

Answer:

Explanation:

Lower the refractive index ,  higher the wave length

Refractive index becoming high reduces the velocity and hence wavelength as frequency remains unchanged.

So refractive index  1.4831 will correspond to wavelength of 450 nm.

For  refractive index is 1.4831 , angle of incidence i , angle of refraction r .

Sin i / Sinr = 1.4831

sin65 / sinr = 1.4831

sir r = sin65 / 1.4831

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= .6111

r = 37.67 degree

For  refractive index is 1.4754 , angle of incidence i

Sin i / Sinr = 1.4754

sin65 / sinr = 1.4754

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6 0
3 years ago
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Answer:

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Explanation:

From the question we are told that:

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Let

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Gravity g=9.8m/s^2

Generally the equation for Gravity at altitude is mathematically given by

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Weight at 6.33 altitude

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 W_a=840.2N

Therefore

 Weight loss=W_s-W_b

 Weight loss=841.82-840.2

 Weight loss=1.6321N

3 0
3 years ago
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