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3 years ago

9

A spring balance used to weigh candy is built with a spring the spring stretches 3.00 cm when a 15 n weight is placed in the pan

. when the weight is replaced with 3 kg of candy, what distance will the spring stretch?
a.7.0 b.4.5 c.3.8 d.5.9

1 answer:

ira [324]3 years ago

6 0

Answer: 5.9cm

Explanation:

15N/3cm=5

Convert 3kg to N (29.42N)

Use the 5 from earlier to divide 29.42/5=5.884

Round to 5.9

5.9cm is the answer

(I don’t know if I solved this the correct way but I got it right. 5.9 is the correct answer)

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What is acceleration produced by a force of 12 newton exerted on an object of mass 3kg

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a=F/m

a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)

a=4m/s^2

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A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

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Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

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Total energy = Kinetic energy + spring potential energy

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So, K.E = Total energy

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K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

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thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

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What is physical measurement? in Science

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3 years ago

Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. w

PSYCHO15rus [73]

1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

$F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N$

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

$F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m$

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s$

And the frequency of oscillation is given by:

$\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz$

3) 2.19 m/s

The velocity at time t of the block is given by:

$v=v_0 cos (\omega t)$

where

$v_0 = 4.4 m/s$ is the initial velocity of the block

$\omega=5.81 rad/s$ is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

$v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s$

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

$a_0 = \omega^2 A$

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

$K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m$

So, the maximum acceleration is

$a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2$

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

$F=mg-kx$

First, we need to find the position x at t=0.36 s, which is given by

$x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m$

And so, the net force is

$F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N$

And the negative sign means the direction of the force is upward.

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3 years ago

Read 2 more answers

An electron of mass 9.11×10-31kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to t

SOVA2 [1]

Answer:

(A) Acceleration will be $240.3846\times 10^{12}m/sec^2$

(b) Time taken will be $1.4\times 10^{-8}sec$

(c) Force will be $2189.9\times 10^{-19}N$

Explanation:

We have given that electron starts from rest so initial velocity u = 0 m/sec

Final velocity $v=2.50\times 10^6m/sec$

Mass of electron $m=9.11\times 10^{-31}kg$

Distance traveled by electron $s=1.30cm =0.013m$

From third equation of motion we know that $v^2=u^2+2as$

(a) So $(2.5\times 10^6)^2=0^2+2\times a\times 0.013$

$a=240.3846\times 10^{12}m/sec^2$

(b) From first equation of motion we know that v = u+at

So $2.50\times 10^6=0+240.3846\times 10^{12}t$

$t=0.014\times 10^{-6}=1.4\times 10^{-8}sec$

(c) From newton's law we know that force

$F=ma=9.11\times 10^{-31}\times 240.3846\times 10^{12}=2189.9\times 10^{-19}N$

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3 years ago