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Lana71 [14]
3 years ago
5

A black stone was discovered to have magnetic properties.

Physics
1 answer:
Jobisdone [24]3 years ago
5 0

Answer:

a) this stone was first called Schorl stone

b)the first modern name for this mineral was Black tourmaline

Explanation:

hope this helps you!

:)

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To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
3 years ago
The electric field due to two point charges is found by: a) finding the stronger field. The net field will just be equal to the
alekssr [168]

Answer:

b)determining the electric field due to each charge and adding them together as vectors.

Explanation:

The electric Field is a vector quantity, in other words it has a magnitude and a direction. On the other hand, the electric field follows the law of superposition. The electric field produced by two elements is equal to the sum of the electric fields produced by each element when the other element is not present. in other words, the total electric field is solved determining the electric field due to each charge and adding them together as vectors.

7 0
3 years ago
A car accelerates from rest for 8.0 s, and reaches a speed of 66 m/s. How far will the
PSYCHO15rus [73]

Answer: C

Explanation:

Find the acceleration using this kinematic equation:

v_f=v_i+at\\66=0+a*8\\a=8.25m/s^2

Now use this kinematic equation to find the displacement:

v_f^2=v_i^2+2a*d\\66^2=0^2+2(8.25)d\\4356=16.5d\\d=264m

3 0
3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306        

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration is given by;

α  = (ωₓ -   ωₐ) /  T₁

α = ((7.272205217  × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

 α  = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α  = 6.152 X 10⁻²⁰ rad/s²

8 0
3 years ago
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