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Anni [7]
2 years 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]2 years 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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A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them
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Answer:

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From the table, the values of force increases with increase in the value of mass.

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Finding the contant of proportionality k;

k=25/5=5

thus M=k(F)...........where M is mass in kg and F is force in newton, then

M=5F

This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

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Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

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And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

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If a river current is 8.0 m/s, and a boat is traveling 10.0 m/s upstream, what is the boat’s speed relative to the riverbank?
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3 0
3 years ago
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a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s
Pepsi [2]

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

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