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3 years ago

If a marble is released from a height of 10 meters how long would it take to hit the ground?

1 answer:

3 0

S = ut + 0.5at^2

10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} 

t^2 ~ 2.04 

t ~ 1.43 seconds

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A solid with flat sides that meet at sharp edges and corners is called

Illusion [34]

1.) Product is a solid
2.) Product has flat slides
3.) Flat sides meet at sharp edges and corners
The product would be a crystal, or it would be a type of product in a crystal structure.

3 0

3 years ago

The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec

marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

$\epsilon = \frac{\sigma}{\sqrt{n} }$

To reduce the standard error to 0.03 s, let the additional number of trials = x

$0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials$

Therefore, the additional trials needed is 48 trials.

6 0

3 years ago

An infinite long straight wire is uniformly charged, the charge density is a. Use Coulomb's law to calculate the electric field

bixtya [17]

Answer:

$\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R$

Explanation:

Since the wire is infinitely long, we will use Gauss' Law:

$\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical surface with height h around the wire. The electric flux through the imaginary surface will be equal to the net charge inside the surface.

In that case, the net charge inside the imaginary surface will be the portion of wire with height h. Then the charge of that portion will be equal to

$Q_{enc} = ah$

The left-hand side of the Gauss' Law is the flux through the imaginary surface. Since we choose our surface as a cylinder, of which we know the area, we do not have to take the surface integral.

$\int\vec{E}d\vec{a} = E2\pi R h$

where R is the radius of the imaginary cylinder.

Finally, Gauss' Law gives

$E2\pi Rh = \frac{ah}{\epsilon_0}\\E = \frac{a}{2\pi \epsilon_0 R}$

The vector expression is

$\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R$

As you can see, the electric field is independent from the height h, since that is merely an imaginary cylinder to apply Gauss' Law. In the end, what matters is the charge density of the wire and the distance from the wire.

4 0

4 years ago

Someone pushes on a rock and measures the net force acting on it is 147 N and acceleration to be 7.5m/s2.What is the weight of t

kogti [31]

Some1 pushes u and then u defend your self

7 0

3 years ago

A vector is given by R 2i + j+3k. Find (a) magnitudes of the x, y, and z components; (b) the ma nitude of R; and (c) the angles

Darya [45]

Answer:

given,

R = 2i + j+3k