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3 years ago

How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?

Physics

2 answers:

Hitman42 [59]3 years ago

3 0

Answer:

C. The block and the sphere would have the same weight.

Explanation:

If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.

ANTONII [103]3 years ago

3 0

C is correct

Sorry if I’m wrong

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In a certain physics experiment, carts run along a frictionless, level track. Cart A is 1 kg and Cart B is 2 kg. Initially cart

timofeeve [1]

Answer: $v=-\frac{11}{3} m/s \left$

Explanation:

Given

Mass of car A $\left ( m_a\right )=1 kg$

Mass of car B $\left ( m_b\right )=2 kg$

velocity of cart A=3 m/s

velocity of cart B=-7 m/s

Conserving momentum

$m_au_a+m_bu_b=\left ( m_a+m_b\right )v$

$v=\frac{m_au_a+m_bu_b}{\left ( m_a+m_b\right )}$

$v=-\frac{11}{3} m/s \left ( i.e. in\ left\right )$

6 0

3 years ago

The magnitude of the electric current is directly proportional to the _____________ of the electric field.

Harlamova29_29 [7]

The magnitude of the electric current is directly proportional to the "Electric Charge" <span>of the electric field.

Hope this helps!</span>

7 0

3 years ago

Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly

AleksAgata [21]

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = $1.989\times 10^{30}\ kg$

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = $147.12\times 10^{9}\ m$

$r_e$ = Radius of Earth = $6.371\times 10^6\ m$

$r_s$ = Radius of Sun = $695.51\times 10^6\ m$

In this system it is assumed that the potential and kinetic energies are conserved

$\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s$

The velocity of Earth would be 613373.65233 m/s

3 0

3 years ago

The sun is more massive than the moon, but the sun is farther from the earth. Which one exerts a greater gravitational force on

love history [14]

Answer:

178.4 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ is the masses of the 2 objects. and R is the distance between them.

From here we can calculate the ratio of gravitational force between the moon and the sun

$\frac{F_s}{F_m} = \frac{G\frac{MM_s}{R_s^2}}{G\frac{MM_m}{R_m^2}}$

We can divide the top and bottom by G and M

$\frac{F_s}{F_m}= \frac{M_s}{R_s^2}:\frac{M_m}{R_m^2}$

$= \frac{M_s}{R_s^2}\frac{R_m^2}{M_m}$

$= \frac{M_s}{M_m}(\frac{R_m}{R_s})^2$

$= \frac{1.99*10^{30}}{7.35*10^{22}}(\frac{3.85*10^8}{1.5*10^{11}})^2$

$= 27074830*6.59*10^{-6} = 178.4$

So the gravitational force of the sun is about 178 times greater than that of the moon to an object on Earth

5 0

3 years ago

A rectangular block measures 4.1cm by 2.8cm by 2.1cm. calculate its volume given you answer to an appropriate number of signific

sattari [20]

Answer:

Volume = 24 cm^3

Explanation:

We recall that the volume of the box is calculated via the formula:

Volume = length * height * width

and that in a product, the number of significant figures for the result should coincide with the number of significant figures of the factor that has the least of them.

In this case, all measures have the same number of significant figures: two. so we calculate the product, and then limit the answer value to exactly two significant figures:

Volume = 4.1 cm * 2.8 cm * 2.1 cm = 24.108 cm^3, which must be rounded to two significant figures as: 24 cm^3

6 0

3 years ago

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