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3 years ago

1.which of the following are true.

Physics

2 answers:

mezya [45]3 years ago

5 0

Answer:

its c i did usatestprep

Explanation:

Ksivusya [100]3 years ago

3 0

B is the correct answer

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While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and

KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement s = final position - initial position

$s=ut+\frac{1}{2}at^{2}$ , where u is the initial velocity, a is the

constant acceleration and t is the time

So we can find the final velocity by using the rule:

final position - initial position = $ut+\frac{1}{2}at^{2}$

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = $(-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}$

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

That means the car is at 8.1625 meters in opposite direction

The position of the car at t = 1.5 s is at -8.1625 meters

4 0

3 years ago

If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?

nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = $\frac{GM}{R^{2} }$

where G is the gravitational constant

Gravitational force of the planet = $\frac{G4M}{(4R)^{2} }$

= $\frac{G4M}{16R^{2} }$

= $\frac{GM}{4R^{2} }$

recall, gravitational force of earth is given as = $\frac{GM}{R^{2} }$

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0

3 years ago

Which of the follow represent a chemical changes

mixer [17]

Answer:

C) steel turning to rust in salt air

Explanation:

The missing options are:

A) ice melting to form liquid water

B) water boiling to form steam

C) steel turning to rust in salt air

D) sugar dissolving into hot coffee

In a chemical change the atoms of the reacting compounds are reordered forming new compounds. In a chemical change, new compounds appear, but in a physical change not.

Then, change of states like ice melting and water boiling are not chemical changes.

During steel rust, components of steel, like iron, are oxidized, that is, reacts with oxygen forming oxides.

The dissolution of sugar into hot coffee is a physical change in which sugar molecules get further apart in the coffee, but they don't change.

5 0

3 years ago

Please answer both questions and not just one. Thanks!

Margarita [4]

It would be d and c hoped i helped!

8 0

3 years ago

In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for

4vir4ik [10]

Answer:

The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is

$k = 0.903$

Explanation:

From the question we are told that

The time constant $\tau = 3$

The potential across the capacitor can be mathematically represented as

$V = V_o (1 - e^{- \tau})$

Where $V_o$ is the voltage of the capacitor when it is fully charged

So at $\tau = 3$

$V = V_o (1 - e^{- 3})$

$V = 0.950213 V_o$

Generally energy stored in a capacitor is mathematically represented as

$E = \frac{1}{2 } * C * V ^2$

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now since capacitance is constant at $\tau = 3$

The energy stored can be evaluated at as

$V^2 = (0.950213 V_o )^2$

$V^2 = 0.903 V_o ^2$

Hence the fraction of the energy stored in an initially uncharged capacitor is

$k = 0.903$

4 0

3 years ago