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3 years ago

Describe a situation where you might have high velocity but low acceleration

1 answer:

6 0

In a level cruise on a passenger airliner, flying in a straight line at a steady 550 miles per hour.
Acceleration is zero.

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Gold has a density of 19.3 g/cm3. What is the mass of a 5 cm3 block of gold?

Tems11 [23]

Do not forget that mass = <span>volume x density
</span>Mass of 1 cm^3 = Density[/tex]
$mass of 2 cm^3 = 19.3 g + 19.3 g = 2*19.3 g$
Then eventually we can find <span>mass of 5 cm^3 : =
</span> $19.3 g + 19.3 g+19.3g+19.3g+19.3g= 5*19.3 g$
So the answer is D
<span>And that's it. I'm sure it will help.</span>

5 0

2 years ago

The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

n200080 [17]

<h2>Answer:g=9.79 $ms^{-2}$ ,A object of mass $m$ at the surface of earth experiences a force $mg$ </h2>

Explanation:

Let $M$ be the mass of earth.

Let $R$ be the radius of earth.

Let $G$ be the universal gravitational constant.

Given,

$M=5.96\times 10^{24}Kg$

$R=6.37\times 10^{6}m$

$G=6.67259 \times 10^{-11}$ $Nm^{2}Kg^{-2}$

Let $g$ be the acceleration due to gravity.

Then, $g=\dfrac{GM}{R^{2}}$

$g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}$

$g=9.79ms^{-2}$

A object of mass $m$ at the surface of earth experiences a force $mg$

3 0

3 years ago

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm

Angelina_Jolie [31]

Answer:

E1 = 2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

= 2996.667N/C

E2 = kQ2/r^2

= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

= 11237.5N/C

The direction are towards the point a

6 0

3 years ago

In an electrochemical cell, the anode is ____.

harkovskaia [24]

A) the electrode at which oxidation takes place

3 0

3 years ago

Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u

tino4ka555 [31]

Answer:

a) force between them is attraction, b) F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

In the exercise they give us the values of the loads

q1 = - 10 mC = -10 10⁻³ C

q2 = 5 mC = 5 10⁻³ C

d = 20 cm = 0.20 m

let's calculate

F = 9 10⁹ $\frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}$

F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0

3 years ago