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Anastasy [175]
3 years ago
14

A soccer player hits a ball with a velocity of 22 m/s at an angle of 36.9 above the horizontal. Air resistance can be ignored. a

. b. c. (3) What are the x and y components of the balls initial velocity? (3) How high does the ball go? (4) How long (time) does it take to get to the maximum height?
Physics
1 answer:
TiliK225 [7]3 years ago
4 0
Answers:

a. Vx=17.6m/s
Vy=13.2m/s

b. 8.90m

c. 1.80s




Vi=22m/s

Theta=36.9deg

Vx=Vi*cos(theta)

Vy=Vi*sin(theta)

h(Max)=Vi^2*sin(theta)^2/2g

t(max height)= Vy/g

Vx=22*cos(36.9)=17.6m/s

Vy=22*sin(36.9)=13.2m/s

h(Max)=22^2*sin(36.9)^2/2(9.8)=8.90m

t(Max height)=17.6/9.8=1.80s
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A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.
ycow [4]

Answer:

σ = 3.402 KPa ,  γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

Find the shearing stress, shearing strain and  shear modulus.​

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

- The shear effect results in a stress in the gelatin block.

- The formulation of stress ( σ ) is given below:

                        σ = F / A

Where,

           A : The surface area of the object that experiences the shear force.

- The top surface have the following dimensions:

          A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

                     σ = 49 / 0.0144

                     σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

                    γ = x / L

                    γ = 10 / 40

                    γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- It is mathematically expressed as a ratio of shear stress  ( σ ) and shear strain ( γ ):

                   G =  σ / γ

                   G = 3.402 / 0.25

                   G = 13.608 KPa

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3 years ago
A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha
Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

x(t) = A\cdot \cos (\omega \cdot t + \phi)

Where:

x(t) - Position of the mass with respect to the equilibrium position, measured in centimeters.

A - Amplitude of the mass-spring system, measured in centimeters.

\omega - Angular frequency, measured in radians per second.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)

Where the maximum acceleration of the system is represented by \omega^{2}\cdot A.

The natural frequency of the mass-spring system is:

\omega = \sqrt{\frac{k}{m} }

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

If k = 12\,\frac{N}{m} and m = 0.40\,kg, the natural frequency is:

\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }

\omega \approx 5.477\,\frac{rad}{s}

Lastly, the maximum acceleration of the system is:

a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)

a_{max} = 359.970\,\frac{cm}{s^{2}}

The maximum acceleration of the system is 359.970 centimeters per square second.

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4 0
3 years ago
A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by
Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0
3 years ago
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irakobra [83]

Answer: 7840N

Explanation:

Given that

Potential energy = ?

Mass of sled = 20-kg

Distance = 40 meters

Acceleration due to gravity = 9.8m/s^2

Recall that potential energy is the energy possessed by a body at rest

i.e potential energy = mass m x acceleration due to gravity g x distance h

P.E = mgh

P.E = 20kg x 9.8m/s^2 x 40m

P.E = 7840N

Thus, the potential energy of the sled is 7840N

5 0
3 years ago
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