ANSWER FOR BRAINLIEST

NikAS [45]

Answer: C

Explanation:

8 0

4 years ago

A baseball is tossed straight up in the air. The table shows the height and velocity of the ball at different times as it moves

Akimi4 [234]

Answer: At that moment, all the baseball's kinetic energy has been converted to potential energy.

Explanation: I took the test

3 0

2 years ago

P14.003A spherical gas-storage tank with an inside diameter of 8.1 m is being constructed to store gas under an internal pressur

Artist 52 [7]

Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

d = inside diameter = 8.1 m

P = internal pressure = 1.26 MPa

σ = 270 MPa

factor of safety = 2

Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?

The allow factor of safety:

$\sigma _{a} =\frac{\sigma }{factor-of-safety} =\frac{270}{2} =135MPa$

The minimun wall thickness:

$t=\frac{Pd}{4\sigma _{a} } =\frac{1.26*8.1}{4*135} =0.0189m$

3 0

3 years ago

Which of the following could be used to convert chemical energy to heat energy?

Mandarinka [93]

Answer:

C

Explanation:

Inside a battery there are chemical reactions. When the battery is powering a flashlight, this chemical energy gives a heat one.

3 0

4 years ago

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Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$

This means that at half the distance, the electric field is four times stronger.

4 0

3 years ago