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mrs_skeptik [129]
3 years ago
8

A block of mass m = 0.13 kg is set against a spring with a spring constant of k1 = 623 N/m which has been compressed by a distan

ce of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 487 N/m.(A) How far d2, in meters, will the second spring compress when the block runs into it?(B) How fast v, in meters per second, will the block be moving when it strikes the second spring?(C) Now assume friction is present on the surface in between the ends of the springs at their equilibrium lengths, and the coefficient of kinetic friction is ?k = 0.5. If the distance between the springs is x = 1 m, how far d2, in meters, will the second spring now compress?
Physics
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

a) 0.11 m

b) 6.9 m/s

c) 0.10 m

Explanation:

We need to apply the conservation of energy theorem:

U_e+K_{o}+U_o=K_f+U_f

because the surface is frictionless all the elastic potential energy is converted in kinetic, so:

\frac{1}{2}*k*x^2+0+0=K_f\\\\K_f=3.1J

the second spring received 3.1J of energy, so:

3.1J=\frac{1}{2}*487N/m*(x_2)^2\\x_2=0.11m

We know the value of the kinetic energy so:

\frac{1}{2}m*(v_2)^2=3.1J\\v_2=\sqrt{\frac{3.1J*2}{0.13kg}}\\v_2=6.9m/s

now if the surface has friction:

3.1J-\µ*m*g*d=K_f\\3.1J-0.63J=K_f\\K_f=2.5J\\\\2.5J=\frac{1}{2}*K*x^2\\x=\sqrt{\frac{2*2.5J}{487}}\\x=0.10m

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Answer:

X is ammeter and Y is voltmeter

Explanation:

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Ammeter is a measuring instrument used to measure the current in a circuit by connecting the meter in series with the circuit.

Ammeter is a measuring instrument used to measure the potential difference between two points by connecting the meter in parallel to the points.

Meter X is an ammeter since it is connected in series and meter Y is a voltmeter because it is connected in parallel.

4 0
3 years ago
How might you use a rope and pulley to move a wagon up a ramp?
sleet_krkn [62]
You could attach the pulley to a secure object on the top of the ramp, and crank the pulley to bring the wagon up said ramp into a loading bay perhaps, or a track. 

Hope I helped. 
7 0
2 years ago
A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

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3 0
3 years ago
g If the x-component of a force vector is 5.69 newtons and its y-component is 8.00 newtons, then what is its magnitude?
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Answer:

F = 9.82 N

Explanation:

given,

Force x-component = 5.69 N

Force y-component = 8 N

magnitude of force = ?

Resultant of force

F = \sqrt{F_x^2 + F_y^2}

F = \sqrt{5.69^2 + 8^2}

F = \sqrt{32.3761 + 64}

F = \sqrt{96.3761}

F = 9.82 N

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