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Alla [95]
4 years ago
12

7) A 0.2 kg mass is dropped from a 10 m high building on top of a spring that is placed vertically. It stops after moving 9.2 me

ters down and stops 0.8 meters above the ground while compressing the spring by 0.5 meters. What is the spring coefficient if air resistance is negligible?
Physics
1 answer:
vivado [14]4 years ago
4 0

Answer:

the spring coefficient is

k=16N/m

Explanation:

Hooks law states that provided the elasticity of a material is not exceeded the extension e is proportional to the applied force

Step one

Analysis of the problem

From analysis of the problem

The mass has a potential energy due to the height it was dropped from, the potential energy is then stored in the spring since it was dropped on the spring which compresses it by 0.5m

Step two

Data

Mass of object m=0.2kg

Height of building =10m

Compression of spring e=0.5m

Spring constant k=?

Step three

According to the principle of energy conservation

mgh=1/2(k*e^2)

Making k subject of formula we have

k=2mgh/e^2

Substituting our data into the expression to get k

Assuming g=9.81m/s

k=2*0.2*10/0.5^2

k=4/0.25

k=16N/m

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The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the
I am Lyosha [343]

Answer: 5.77\ rps

Explanation:

Given

Initial angular velocity is \omega_i=8\ rps

rate of reduction \alpha=0.9 rev/s^2

after 17 revolution i.e. \theta =17\ rev

using \Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta

Insert the values

\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps

5 0
3 years ago
What must occur within a magnetic material before it exhibts magnetic properties?
erma4kov [3.2K]

Answer:

Option C, The magnetic domains inside the material must be aligned

Explanation:

Magnetic domains determine the magnetic behavior of any substance. Most of the magnetic substances have disoriented magnetic domains due to which they do not behave as magnets until unless an external force is applied to change the orientation of their domains and align them properly.

Hence, option C is correct

3 0
3 years ago
1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of
Masja [62]

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = \sqrt{3^{2} +2^{2}  } = \sqrt{13} = 3.61cm = 0.036m

r₂ = \sqrt{4^{2} + 3^{2}  } = \sqrt{25} = 5cm = 0.05m

electric potential V = \frac{kq}{r}

change in potential ΔV = V_{1} - V_{2}

ΔV = \frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} } , where q_{1} = q_{2}=2.00μC

ΔV = 2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × (\frac{1}{0.036} - \frac{1}{0.05} )

ΔV= 2.789×10⁵

\frac{1}{2}mv^{2} = ΔV × q₃

\frac{1}{2} ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

3 0
3 years ago
If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin
IceJOKER [234]

Answer:

1.68 \frac{m}{s^2}

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\

Substituting the given values in above equation, we get -

V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}

Now we will determine the radius of curvature for the curve shown in the attached image

Y = 16 - \frac{1}{625} X^2\\

Differentiating on both the sides, we get -

\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} =  -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}  }{\frac{d^2y}{d^2x}} \\Curve = 312.5meter

Acceleration on curved path

a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}

Final acceleration

a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}

5 0
3 years ago
After a nucleus with 85 protons undergoes alpha decay, it has<br><br> ? protons.
Valentin [98]

Answer:

After a nucleus with 85 protons undergoes alpha decay, it has  83 protons.

Explanation:

In an alpha particle there are two protons

In the given substance's nucleus, there are total of 85 protons

After the decay, the proton number reduce

The current proton number after decay is

85 -2 = 83

After a nucleus with 85 protons undergoes alpha decay, it has  83 protons.

5 0
3 years ago
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