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4 years ago

What is the density of 10 cm^3 volume of water that has a mass of 10g

Physics

1 answer:

victus00 [196]4 years ago

5 0

Answer:

Density of water is 1 g/cm³.

Explanation:

Given:

Mass of given volume of water is, $m=10\ g$

Volume of the given amount of water is, $V=10\ cm^3$

We know that, density of a quantity is given as the ratio of its mass to that of its volume.

Here, in order to calculate density of water, we need to divide the mass of water by its volume. Therefore,

$Density=\frac{m}{V}$

Plug in 10 g for 'm', 10 cm³ for 'V and solve for density. This gives,

$Density=\frac{10}{10}\ g/cm^3\\\\Density=1\ g/cm^3$

Therefore, the density of water of mass 10 g and volume 10 cm³ is 1 g/cm³.

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Answer:

600N(downwards)

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600N(downwards)

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3 years ago

A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe

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Answer:

$56.5\ \text{s}$

$21.13\ \text{m/s}$

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}$

Time the car is at constant velocity is 39 s

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Total time the car is in motion is $12.5+39+5=56.5\ \text{s}$

Distance traveled

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$s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}$

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The total displacement of the car is $156.25+975+62.5=1193.75\ \text{m}$

Average velocity is given by

$\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}$

The average velocity of the car is $21.13\ \text{m/s}$ .

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3 years ago

In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help

Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

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3 years ago

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