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Anni [7]
1 year ago
11

An object, whose mass is 0.660 kg, is attached to a spring with a force constant of 132 N/m. The object rests upon a frictionles

s, horizontal surface (shown in the figure below).
An object labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the object is displaced by a distance A to the right of its equilibrium position.
The object is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a)
At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
N
(b)
At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
Toward the equilibrium position (i.e., to the left in the figure).
Away from the equilibrium position (i.e., to the right in the figure).
The direction is not defined (i.e., the acceleration is zero).
You cannot tell without more information.

Physics
1 answer:
Arada [10]1 year ago
7 0

Answer:

See below

Explanation:

a)  Spring force at release =  k * d = 132 N/m * .120 m = 15.84 N

b) F = ma

   15.84 = (.660 kg)(a)     a = 24 m/s^2

c) Toward the left ....the object is accelerated to the left

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An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting
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Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat  * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f  - T ico) + m ice * L + m ice * c wat  * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat  * T f  = m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L

T f  = (m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C +  106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C +  106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

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5 0
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An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east. Immediate
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Answer:

6.25 m/s

Explanation:

mass of man (m1) = 80 kg

mass of boy (m2) = 20 kg

mass of man and boy after collision (m12)= 20 + 80 = 100 kg

velocity of man and boy after collision (v) = 2.5 m/s

angle θ = 60 °

How fast was the boy moving just before the collision ?

  • From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
  • M₁₂ =  total momentum after collision = m12 x v = 100 x 2.5 = 250
  • Mboy = momentum of the boy before collision = m2 x Velocity of boy
  • Mman = momentum of the man before collision = m1 x velocity of man  
  • from the triangle, cos θ = \frac{Mboy}{M₁₂}

        cos 60 = \frac{Mboy}{250}

        Mboy = 250 x cos 60 = 125

  • recall that momentum of the boy (Mboy) also = m2 x Velocity of boy

        therefore

        125 = 20 x velocity of boy

         velocity of boy = 125 / 20 = 6.25 m/s

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