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liraira [26]
3 years ago
7

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

rom the horizontal. Assume no friction. Find the X component for the force. Find the Y component for the force. if the dogs mass is 25 kg find the force of gravity acting on the dog. what is the magnitude of the normal force?

Physics
1 answer:
Delvig [45]3 years ago
6 0

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

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

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N

Therefore, the normal force acting on the dog is 210 N.

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Answer:

Yes.

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2 years ago
In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =
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Answer:

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

Area = pi * (22 * 10^-^3)/4

Area = 3.801 * 10^(-4)    meter squared

Stress = force / area

Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa     OR       0.197 GPa

Modulus of elasticity = stress / strain

200 = 0.19732 / Strain

Strain = 0.0009866     (longitudinal)

(a) Strain = change in length / initial length

0.0009866 = Elongation of rod / 200

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(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

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Lateral strain = Change in diameter / original diameter

0.000296 = Change in diameter / 22

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4 0
3 years ago
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Which of the following methods has led to the earliest discoveries of massive planets orbiting near their parent stars a. detect
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c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

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Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

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\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

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t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

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         x (α x³ + β) = K_{f} - K₀

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Let's calculate

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        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
2 years ago
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