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3 years ago

Which would usually be studied by a geologist

Physics

2 answers:

AnnyKZ [126]3 years ago

5 0

Geologists study Earth processes

xxTIMURxx [149]3 years ago

4 0

Has to do with earth physical structure
Hope this helps :)

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A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0

3 years ago

Only 3 questions plz answer

BigorU [14]

3. D 4. C 5. A. I think

8 0

3 years ago

Read 2 more answers

There are (one can say) three coequal theories of motion for a single particle: Newton's second law, stating that the total forc

PtichkaEL [24]

Answer:

vf = 14.2176 m/s

Explanation:

Given

m = 4 Kg

viy = 7.00 ĵ m/s

Fx = 11.0 î N

t = 4.5 s

vf = ?

Using the Impulse - Momentum Theorem, we have

F*Δt = m*Δv ⇒ F*Δt = m*(vf - vi)

⇒ vf = (F*Δt + m*vi) / m

For x-component

⇒ vfx = (Fx*Δt + m*vix) / m = (11 N*4.5 s + 4 Kg*0 m/s) / (4 Kg)

⇒ vfx = 12.375 î m/s

For y-component

⇒ vfy = (Fy*Δt + m*viy) / m = (0 N*4.5 s + 4 Kg*7 m/s) / (4 Kg)

⇒ vfy = 7 ĵ m/s

Finally:

vf = √(vfx² + vfy²)

⇒ vf = √((12.375 m/s)² + (7 m/s)²)

⇒ vf = 14.2176 m/s

8 0

3 years ago

On which one of the surfaces listed below will a toy car released with the same initial speed travel the farthest? *

irinina [24]

Are you gonna list the surfaces or nah

5 0

2 years ago

What is the direction of a vector with an x component of -12 m and a y component of 21 m?

Svet_ta [14]

Answer:

$Direction= 119.74^\circ$

Explanation:

Displacement Vector

The displacement, as every vector, has a magnitude r and a direction angle θ measured from the positive x-axis.

If we know the x-y components of the displacement, the magnitude and angle can be calculated by the equations:

$r=\sqrt{x^2+y^2}$

$\displaystyle \tan\theta=\frac{y}{x}$

The coordinates of the given vector are x=-12 m, y=21 m, thus:

$\displaystyle \tan\theta=\frac{21}{-12}=-1.75$

$\theta=tan^{-1}(-1.75)=-60.26^\circ$

Since the vector lies in the second quadrant, we add 180° to find the correct direction:

$\boxed{Direction=-60.26^\circ+180^\circ= 119.74^\circ}$

4 0

3 years ago