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Cerrena [4.2K]
3 years ago
6

Use the factorization to find the values of x for

Physics
2 answers:
TiliK225 [7]3 years ago
8 0

Answer:

Nah son

Explanation:

olga2289 [7]3 years ago
3 0

Answer(x – 1)(x + 1)(x + 5) = 0

x – 1 = 0, so x = 1

x + 1 = 0, so x = -1

x + 5 = 0, so x =  -5

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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that
Katyanochek1 [597]

To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.

The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:

F = Eq

F is the electric force, E is the electric field, and q is the charge

Let us adjust the equation for only the horizontal components of the above quantities:

Fx = (Ex)(q)

Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.

Given values:

F = 6.57×10⁻²N

q = 6.80×10³C

Plug in these values and solve for Ex:

6.57x10⁻² = Ex(6.80×10³)

<u>Ex = 9.66×10⁻⁶N/C</u>

<u />

Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.

We know the weight of the sphere is given by:

W = mg

W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.

We also know this equation:

F = Eq

Let us adjust for the vertical components:

Fy = (Ey)(q)

Set Fy equal to W and we get:

(Ey)(q) = mg

Given values:

q = 6.80×10³C

m = 0.018kg

g = 9.81m/s²

Plug in the values and solve for Ey:

(Ey)(6.80×10³) = 0.018(9.81)

<u>Ey = 2.60×10⁻⁵N/C</u>

<u />

Let's now use the Pythagorean theorem to find the total magnitude of the electric field:

E = \sqrt{Ex^{2}+Ey^{2}}

E = 2.77×10⁻⁵N/C

The direction of the electric field is given by:

θ = tan⁻¹(Ey/Ex)

θ = 20.4° off the horizontal

4 0
3 years ago
A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.
Olegator [25]

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

v₀ = 0 m/s

a = 10 m/s²

Find: t

Δy = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (10 m/s²) t²

t = 3.2 s

b) Given in the x direction:

v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²

Δx = 38 m

6 0
2 years ago
What name is given to the wall of water that makes landfall just ahead of a hurricane?
andreev551 [17]
Pretty sure its a. storm surge 
8 0
3 years ago
(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this soleno
Paul [167]

Answer:

a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H

Explanation:

a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.

Also, Li = Ф where L = inductance of solenoid.

So, Li = NAB

= NA(μ₀ni)

= NA(μ₀Ni/l)

Li = μ₀AN²i/l

dividing both sides by i, we have

So, L = μ₀AN²/l

b. The self- inductance, L = μ₀AN²/l where

A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m

So, L = μ₀AN²/l

L = μ₀πd²N²/4l

L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)

L = 11,103.3 × 10⁻¹¹ H

L = 1.11033 × 10⁻⁷ H

L ≅ 1.11 × 10⁻⁷ H

6 0
3 years ago
Ask Your Teacher In a choir practice room, two parallel walls are 4.00 m apart. The singers stand against the north wall. The or
igomit [66]

Answer: 4m

Explanation:

Since the angle of incidence of a plane mirror can be anything from 0 to 90°

Assuming that the place is a perfectly square 4×4m room

The incident ray would be 45° for the choir(object) at a 4m distance, this is still within the range of values.

We do not forget also, that the focal length of a plane mirror is infinity, the organist would in fact see farther than 4m if need be. And wider

5 0
3 years ago
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