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3 years ago

What type of stress can cause an anticline? A. Compression B. force C. shear

Physics

2 answers:

Andrei [34K]3 years ago

4 0

The stress that can cause on the anticline , is the Compression

Lorico [155]3 years ago

3 0

I think it's compression, but I'm not sure.

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A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg,

solmaris [256]

There are two ways to find energy. Energy=F*d=mv^2. We can use this relationship to find v:
$Fd=0.5mv^{2} \\ 80*4=0.5*10v^{2} \\ 64=v^{2} \\ v=8 m/s$

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3 years ago

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

alukav5142 [94]

Answer:

0.00016 kg

Explanation:

Given:

Power = P = 1.2 × 10⁹ Watts

Power = work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W

Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²

= 0.00016 kg

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3 years ago

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

Svet_ta [14]

Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

E $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3

so if R become 2R

E becomes 1/8 of the initial electric field.

8 0

2 years ago

Read 2 more answers

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

As a truck rounds a curve, a box in the bed of the truck slides to the side farthest from the center of the curve. This movement

Bumek [7]

Because I don’t have choices I’m not sure what to chose from or what the subject is about but I can tell you that the box moves from the result of Inertia, and is kenetic energy

3 0

2 years ago