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tigry1 [53]
2 years ago
10

A 20-cm long spring is attached to the wall. When pulled horizontally with a force of 100N, the spring stretches to a length of

22cm. What is the value of the constant? A.) The same spring is used in a tug of war. Two people pull on the ends, each with a force of 100N. How long is the stretched string? B.) The same spring is now suspended from a hook and a 10.2kg block is attached to the bottom end. How long is the stretched spring?
Physics
1 answer:
suter [353]2 years ago
3 0

Answer:

Explanation:

Part 0

All the spring moves is 2 cm

x = 2 cm * [1 m / 100 cm ]

x = 0.020 meters

F = k*d

100N = k * 0.02 m

100 N / 0.02 = k

5000 N / m

Part A

The spring feels a force of 100 N - - 100N = 200 N because each person is pulling in the opposite direction.

F = k * x

200N = 5000 N/m * d

200 / 5000 = d

d = 0.04 meters.

Part B

10.2 kg must be converted to a force as experienced here on earth.

F = m * g

g = 9.81

m = 10.2

F = 10.2 * 9.81

F = 100.06 N

F = k * d

100.06 = 5000 * d

d = 100.06 / 5000

d = 0.02 meters.

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A person is standing outdoors in the shade where the temperature is 35 °C.
Free_Kalibri [48]

Answer:

(a) Eₐ = 6.36 J/s

(b) Eₐ = 4.64 J/s

Explanation:

Stefan-Boltzmann law: States that the total energy per second radiated or absorbed by a black body is directly proportional to the absolute temperature.

Using, Stefan-Boltzmann equation

Eₐ =eσAT⁴ ................ Equation 1

where Eₐ = Radiant energy absorbed per seconds, e = emissivity, σ = stefan - boltzman constant, A = Surface area. and T = temperature in kelvin

(a) Where e = 0.89, σ = 5.67 ×10⁻⁸ watt/m²/K⁴, A = 140 cm² = 140 cm²(m²/10000cm²) = 0.014 m², T = 35 °C = (35 + 273) K = 308 K.

Applying these values in equation 1 above,

Eₐ = 0.89 × 5.67 ×10⁻⁸ × 0.014 × (308)⁴

Eₐ =6.36 J/s

(b) when e = 0.65,

∴ Eₐ = 0.65 × 5.67 × 10⁻⁸ × 0.014 × (308)⁴

 Eₐ = 4.64 J/s

6 0
2 years ago
True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then th
nlexa [21]
I think that this is false but I am not sure
5 0
1 year ago
Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. At lef
Sladkaya [172]

Question:

Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. In which direction should an arrow at position 1 point?

left

right

up

down

Answer:

The correct answer is

Left

Explanation:

Magnetic circuits describe the path of a magnetic flux. In the same way electricity follows a complete closed circuit, the path of a magnetic flux is also a complete and closed circuit which leaves from the N pole, migrates through the air  and reenters the magnet through the S pole through which it passes back into the magnet to come to the N pole again.

As such the magnetic field lines emanate from the N pole which is on he right to the S pole which is on the left. Hence the arrow should point in the left direction.

3 0
2 years ago
Read 2 more answers
If the diameter of a radar dish is doubled, what happens to its resolving power, assuming that all other factors remain unchange
Tomtit [17]

Answer:

e. Doubles.

Explanation:

Resolving power is given by the formula as follows :

\dfrac{1}{d\theta}=\dfrac{D}{1.22\lambda}

Here, d\theta is the angle subtended by two distant objects

D is diameter of the telescope

Here, the diameter of a radar dish is doubled, assuming all other factors remain unchanged, then the resolving power gets doubled. Hence, the correct option is (e).

7 0
2 years ago
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
7nadin3 [17]

Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

           \varepsilon _{0} is permittivity of free space = 8.85 x 10^{-12} C^{2}/N-m^{2}

           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

                                        = 1800 m

∴ Total energy stored in the capacitor

U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad

U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800

U=10.322 J

4 0
2 years ago
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