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2 years ago

Electrical force is much weaker than the force of gravity true or false.

Physics

2 answers:

docker41 [41]2 years ago

3 0

Answer:

false

Explanation:

The electrical force is much weaker than the force of gravity. False. Thomas Edison argued that our electrical system should use direct current because he felt that alternating current was dangerous.

Nookie1986 [14]2 years ago

3 0

Answer:

False

Explanation:

electrical force is billions and trillions and trillions times stronger than gravity. Gravity is a weak force compared to electrical force.

brainliest would be appreciated?

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A rock hits a window and stops in 0.15 seconds. The net force on the rock is 58N during the collision. What is the magnitude of

nlexa [21]

Answer:

The change in momentum is $\Delta p = 0.7 \ kg\cdot m \cdot s^{-1}$

Explanation:

From the question we are told that

The time taken for the stone to stop is $\Delta t = 0.15 \ seconds$

The net force on the rock is $F = 58 \ N$

The impulse of the rock can be mathematically represented as

$I = F * \Delta t$

Substituting values

$I = 58 * 0.15$

$I = 0.7\ kg * m * s^{-1}$

Now impulse is defined as the rate at which momentum change

Hence the change in momentum $\Delta p$ of the rock is equal to the impulse of the rock

$\Delta p = I = 0.7 \ kg\cdot m \cdot s^{-1}$

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3 years ago

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Nadusha1986 [10]

Answer:

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Explanation:

3 0

2 years ago

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We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap

marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B= $3.7\times 10^{-2} T$

Current,I=4.6 A

Diameter of wire,d=0.5 mm= $0.5\times 10^{-3} m$

Radius of wire,r= $\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm= $1\times 10^{-2} m$

$1 cm=10^{-2} m$

Resistivity of copper, $\rho=1.68\times 10^{-8}\Omega m$

We know that

$R=\frac{\rho l}{A}$

Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire= $\frac{l}{2\pi r'}$

Number of turns of wire= $\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the solenoid,N=799

Magnetic field in solenoid,B= $\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

$n=\frac{N}{L}$

$L=\frac{N}{n}$

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

Length of solenoid=12.5 cm

1m=100 cm

8 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

3 years ago

Explain how do you determine the intensity of earthquake

Ivanshal [37]

Answer:

The Richter scale measures the largest wiggle (amplitude) on the recording, but other magnitude scales measure different parts of the earthquake. The USGS currently reports earthquake magnitudes using the Moment Magnitude scale, though many other magnitudes are calculated for research and comparison purposes.

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3 years ago

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Electrical force is much weaker than the force of gravity true or false. ​

Electrical force is much weaker than the force of gravity true or false.