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NARA [144]
3 years ago
11

A mass of 1.03 kilograms is placed on a horizontal frictionless surface against an uncompressed spring with spring constant 861.

1 N/m. The mass is pushed against the spring until the spring is compressed a distance 0.36 m and then released. How high (vertically) in m does the mass rise from the original height before it stops (momentarily).
Physics
1 answer:
Lelechka [254]3 years ago
8 0

Answer:

5.522 m

Explanation:

Data provided:

Mass, m = 1.03 kg

spring constant, k = 861.1 N/m

Distance by which the spring is compressed, x = 0.36

Thus,

the energy stored in the spring = \frac{1}{2}kx^2

on substituting the values, we get

the energy stored in the spring =  \frac{1}{2}\times861.1\times0.36^2

now,

by the conservation of energy, we have

Potential energy gained by the mass =  Energy gained by the spring

or

mgh = \frac{1}{2}\times861.1\times0.36^2

where,

g is the acceleration due to the gravity

h is the maximum height reached by the mass before falling

on substituting the values in the above relation, we get

1.03 × 9.81 × h = \frac{1}{2}\times861.1\times0.36^2

or

h = 5.522 m

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2 years ago
An electric heater is operated by applying a potential difference of 78.0 V across a nichrome wire of total resistance 7.00 Ω. a
aleksley [76]

PART A)

Here we know that

potential difference across the wire is

V = 78 volts

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R = 7 ohm

now by ohm's law

V = iR

78 = i(7 ohm)

i = 11.14 A

Part b)

Power rating is defined as rate of electrical energy

it is defined as

P = iV

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8 0
3 years ago
With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as
Nonamiya [84]

Answer:

|Δf| = 37.3 kHz

Explanation:

given,

peak velocity = 4 m/s

speed of the sound = 1500 m/s

frequency = 7 MHz

v = C\dfrac{\pm \dlta f}{2 f_0}

\delta f = \pm 2 f_0 (\dfrac{V}{C})

\delta f = \pm 2\times 7 (\dfrac{4}{1500})

           =\pm 0.0373 MHz

           = 37.3 kHz

|Δf| = 37.3 kHz

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4 0
3 years ago
The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t
DiKsa [7]
I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
E_p=\frac{1}{2}kx^2
The kinetic energy of two carts is:
E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
So we have:
E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
Momentum also has to be conserved:
m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}
Now we go back to the momentum equation:
v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}

8 0
3 years ago
Pls help I will mark brainliest
DiKsa [7]

Remember that two of the same charge repel each other.

These two balls are repelling each other.  So they must both have the <em>SAME charge</em>. (last choice on the list)

5 0
3 years ago
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