I need to make a project that represents united together any ideas?

Which type of rock is made of tiny bits of animal shells? A. gabbro B. coquina C. clastic sandstone D. limestone

marysya [2.9K]

Hello,

The answer is <span>C. clastic sandstone

Hope this helps</span>

8 0

3 years ago

A 28-kg beginning roller skater is standing in front of a wall. By pushing against the wall, she is propelled backward with a ve

Scorpion4ik [409]

Answer:

33.6 Ns backward.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = change in momentum

I = mΔv................................. Equation 1

Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.

Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)

Substituting into equation 1

I = 28(-1.2)

I = -33.6 Ns

Thus the impulse = 33.6 Ns backward.

3 0

4 years ago

How man significant figures does the following value gave 43.023

Varvara68 [4.7K]

Answer:

It has 5 I hope this helps you

7 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

9- When the moon comes in an orbit higher from earth, ......... solar eclipse are formed.

OleMash [197]

C partial solar eclipse are formed

5 0

3 years ago