All categories

3 years ago

How to get infinite thanks could you show me please ☺️

Physics

1 answer:

murzikaleks [220]3 years ago

7 0

Answer: divide by zero, or square root of a negative

Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.

A couple of ways to get that...

You try to divide by zero. In other words, a problem that asks you to perform something like this: 5/0= or 23/(4-4)= Such a problem will give you an error on a calculator because the answer is infinity or doesn't exist.
Another way is to try to get the square root of a negative number. That answer doesn't exist as a real number, so $\sqrt{-4\\}$ will also give you an error on a calculator.

You might be interested in

Electricity is the movement of _____ from one atom to another along a conductor.

prohojiy [21]

The sentence can be completed as follows:

"Electricity is the movement of charges from one atom to another along a conductor."

The movement of the the charges is due to the presence of a potential difference across the conductor, which pushes the charges. More specifically, in normal conductors, the carriers of electricity are the electrons. Therefore we can rewrite the sentence as

"Electricity is the movement of electrons from one atom to another along a conductor."

4 0

3 years ago

Can someone help with thsi? i will give brainliest

kogti [31]

Answer:

40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.

Explanation:

4 0

3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

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

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$