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Alla [95]
3 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]3 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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D. 'g' vanishes at centre of
guapka [62]

Answer:

a. 0.8 cm

Explanation:

The distance of the object from the lens, u = 1 cm

The magnification of the lens, m = 5

The focus of a lens formula is given as follows;

f = \dfrac{1}{\dfrac{1}{v} + \dfrac{1}{v}  }

The magnification of the lens, m = -v/u

Where;

v = The distance of the image from the lens

Therefore, we have;

v = m × u

∴ v = 5 × 1 cm = 5 cm (on the other side of the lens)

From which we get;

f = \dfrac{1}{\dfrac{1}{1} + \dfrac{1}{5}  } = \dfrac{5}{6}  \approx 0.8

The focal length ≈ 0.8 cm

6 0
3 years ago
Point X is midway between the charges. In what section of the line will there be a point where the resultant electric field is z
Fiesta28 [93]

Answer:

I believe the answer is in fact section (VW) on the line where the electric field result will be zero.

Explanation:

The direction of the electric field due to a positive charge is away from it and the direction of the electric field due to a negative one is towards it.

3 0
3 years ago
A vertical steel beam in a building supports a load of 6.0×10⁴. If the length of the beam is 4.0m and it's cross-sectional area
Furkat [3]

Answer:

DL = 1.5*10^-4[m]

Explanation:

First we will determine the initial values of the problem, in this way we have:

F = 60000[N]

L = 4 [m]

A = 0.008 [m^2]

DL = distance of the beam compressed along its length [m]

With the following equation we can find DL

\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m]

Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.

A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D =  \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m]

3 0
4 years ago
A mass of 0.34 kg is fixed to the end of a 1.4 m long string that is fixed at the other end. Initially at rest, he mass is made
frozen [14]

At time t seconds, the mass has angular speed

\omega = \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t

and hence linear speed

v = (1.4\,\mathrm m) \omega = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t

After 8 s, its linear speed is

v = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) (8\,\mathrm s) = 37.072 \dfrac{\rm m}{\rm s} \approx 37 \dfrac{\rm m}{\rm s}

and has centripetal acceleration with magnitude

a = \dfrac{v^2}{1.4\,\rm m} \approx 981.667\dfrac{\rm m}{\mathrm s^2} \approx 980 \dfrac{\rm m}{\mathrm s^2}

To maintain this linear speed, by Newton's second law the required centripetal force should have magnitude

F = (0.34\,\mathrm{kg}) a \approx 333.767\,\mathrm N \approx \boxed{330 \,\mathrm N}

5 0
2 years ago
The primary difference between a barometer and a manometer is
Nitella [24]

Answer:

a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.

Explanation:

A barometer measures air pressure at any locality with sea level as the reference.

However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure

The algebraic sum of these two values gives the absolute pressure.

4 0
3 years ago
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