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cupoosta [38]
3 years ago
9

It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

e’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective force constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85-kg team?
Physics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

(a)spring effective force constant =1568N/cm

(b) Yes

Explanation:

Hooke's law is represented mathematically as, F=ke

where

  • F is force applied to elastic material,
  • k= spring constant
  • e= extension

(a).After the maximum load is exceeded, Hooke's law doesn't apply anymore. from the problem, its stated that a maximum load of 120kg will cause an extension of 0.75cm, we will use this to determine the spring constant.

F=mg, g=9.8m/s^{2}

F= 120*9.8 =1176N

From Hooke's law, k=\frac{F}{e}

k=\frac{1176}{0.75}

k=1568N/cm

(b). The players who stands on the scale causes a 0.48cm extension.

    e= 0.48cm, k= 1568N/cm

F=Ke

F= 1568*0.48

F= 752.64N

To calculate the mass of the player we divide this force by g=9.8m/s^{2}

F=mg

m=F/g

m= \frac{ 752.64}{9.8} \\\\m=76.8kg

since the rugby team is an under 85kg team, this player with 76.8kg mass is eligible

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icang [17]

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

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A particle moves along the x axis. It is intially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.32
Nataliya [291]

Answer:

The position of the particle is -2.34 m.

Explanation:

Hi there!

The equation of position of a particle moving in a straight line with constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the particle at a time t:

x0 = initial position.

v0 = initial velocity.

t = time

a = acceleration

We have the following information:

x0 = 0.270 m

v0 = 0.140 m/s

a = -0.320 m/s²

t = 4.50 s  (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).

Then, we have all the needed data to calculate the position of the particle:

x = x0 + v0 · t + 1/2 · a · t²

x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²

x = -2.34 m

The position of the particle is -2.34 m.

6 0
3 years ago
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger
Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

i1•r2 = i2•r1

r2 = i2•r1 / i2

r2 = 20×3.4/12

r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

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