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Anton [14]
3 years ago
15

a load of 45 N attached to a spring that is hanging vertically stretches the spring 0.14 m. What is the spring constant?

Physics
2 answers:
vredina [299]3 years ago
4 0

Answer:

Spring constant, k = 321.42 N/m

Explanation:

Force acting on spring, F = 45 N

Displacement caused due to stretching, x = 0.14 m

Using Hooke's law :

F=-kx

Where

k is the spring constant

k=\dfrac{F}{x}

k=\dfrac{45\ N}{0.14\ m}

k = 321.42 N/m

Hence, this is the required solution.

zhenek [66]3 years ago
3 0
According to Hooke's law, Force = spring constant x displacement of the spring. Spring constant = Force/displacement in spring = 45/0.14 = 321.42 N/m. Hope this helps!
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A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
2 years ago
If a sled has a mass of 4kg what is the force of gravity on the sled?
ale4655 [162]

Answer:

Explanation:

Gravity pulls everything down at the same rate of 9.8 m/s/s. If you're looking for the normal force, which is the same as the weight of the object, we'll find that, just in case.

w = mg which says that the normal force/weight of an object is equal to its mass times the pull of gravity:

w = 4.0(9.8) so

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7 0
3 years ago
A 45.0-N force pushes a cart 12.5 meters down a hallway. What is the work done on the cart?
makvit [3.9K]

Answer:

562.5J

Explanation:

The following were obtained from the question:

F = 45N

d = 12.5m

w =?

The work done can be achieved by using

w = F x d

w = 45 x 12.5

w = 562.5J

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3 years ago
What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are
nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is 15.58 g/cm^3.

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = 19.32g/cm^3

The density of silver = 10.1g/cm^3

The density of copper =8.8g/cm^3

Volume=\frac{Mass}{Density}

Volume of the gold in the mixture = V_1=\frac{18x}{19.32 g/cm^3}

Volume of the silver in the mixture = V_2=\frac{5x}{10.1 g/cm^3}

Volume of the copper in the mixture = V_3=\frac{1x}{8.8 g/cm^3}

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3

8 0
3 years ago
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