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3 years ago

Maya kicks a soccer ball 40 N towards east. At the exact same time, Casey kicks

Physics

1 answer:

Akimi4 [234]3 years ago

3 0

Answer:

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Explanation:

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How many volts does it take to push 2 amperes of current through 20 ohms of resistance?

vodomira [7]

As we know V=IR

So, R= 20 ohms, I= 2 A

Therefore, V= 2*20 = 40 V

Electric Voltage

The difference in electric potential between two points, also known as voltage, electric potential difference, electric pressure, or electric tension, determines how much labour is required to move a test charge between the two sites in a static electric field. Volt is the name of the derived unit for voltage (potential difference) in the International System of Units. 1 volt equals 1 joule (of work) for 1 coulomb, is how to work per unit charge is stated in SI units (of charge). Power and current were utilized in the previous SI definition for the volt, and the quantum Hall and Josephson effect was incorporated in 1990. As of 2019 fundamental physical constants have been employed in the definition of all SI units and derived units.

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2 years ago

Why doesn't the earth block out the light between the sun and the moon during a normal month

almond37 [142]

Answer:

Explanation:

6 0

3 years ago

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are 0.50 L, 2.

aleksklad [387]

Answer:

13.524 N

Explanation:

Volume and densities are given as:

ρ1 = 2.6 g/cm³ => 2600 kg/m³ ; V1 = 0.50 L => 0.5 x 10^-3 m³

ρ2 = 1.0 g/cm³ => 1000 kg/m³ ; V2= 0.25 L => 0.25 x 10^-3 m³

ρ3 = 0.7 g/cm³ => 700 kg/m³ ; V3 = 0.4 L => 0.4 x 10^-3 m³

Next is to calculate force exerted on the bottom of the container due to these liquids:

F= ρ1V1g + ρ2 V2 g+ ρ 3 V3g

where ,

ρ= density

V= volume

g= 9.8m/s²

F= g( 2600 x 0.5 x 10^-3 + 1000 x 0.25 x 10^-3 + 700 x 0.4 x 10^-3)

F= 9.8 (1.38)

F= 13.524 N

Therefore, the force on the bottom of the container due to these liquids is 13.524 N

6 0

4 years ago

Gibbons move through the trees by swinging from successive handholds, as we have seen. To increase their speed, gibbons may brin

Sedbober [7]

By bring their legs close to their bodies, they are decreasing the length of the pendulum which help them move more quickly.

A pendulum is nothing but a body suspended from a fixed point so that it can swing back and forth under the influence of gravity.

Here in this case Gibbons are bringing their legs close to their bodies and reducing the length of the pendulum. Since as the length of the pendulum increases the speed of the movement will be reduced. By bringing their legs close to their bodies they are reducing the length and in turn their speed increase and they move quickly.

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5 0

2 years ago

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

$v = \frac{{2\pi L}}{T}$

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

$v = \frac{{2\pi L}}{T}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{2\pi }}{T}$ is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute $v = \frac{{2\pi L}}{T}$ in the force expression ${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$ as follows:

$\begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{4{\pi ^2}}}{{{T^2}}}$ not affect the force value for six satellites.Therefore, we can write the expression of ${F_c}$ given as follows:

${F_c} = kmL$

Here, k refers to constant value and equal to $\frac{{4{\pi ^2}}}{{{T^2}}}$

${F_A} = k{m_A}{L_A}$

Substitute 200 kg for ${m_A}$ and 5000 m for LA in the expression ${F_A} = k{m_A}{L_A}$

$\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite B from their rocket is given as follows: ${F_B} = k{m_B}{L_B}$

Substitute 400 kg for ${m_B}$ and 2500 m for in the expression ${F_B} = k{m_B}{L_B}$

$\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite C from their rocket is given as follows: ${F_C} = k{m_C}{L_C}$

Substitute 100 kg for ${m_C}$ and 2500 m for in the above expression ${F_C} = k{m_C}{L_C}$

$\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite D from their rocket is given as follows: ${F_D} = k{m_D}{L_D}$

Substitute 100 kg for ${m_D}$ and 10000 m for ${L_D}$ in the expression ${F_D} = k{m_D}{L_D}$

$\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite E from their rocket is given as follows: ${F_E} = k{m_E}{L_E}$

Substitute 800 kg for ${m_E}$ and 5000 m for ${L_E}$ in the expression ${F_E} = k{m_E}{L_E}$

$\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite F from their rocket is given as follows: ${F_F} = k{m_F}{L_F}$

Substitute 300 kg for ${m_F}$ and 7500 m for ${L_F}$ in the expression ${F_F} = k{m_F}{L_F}$

$\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The value of forces obtained for the six-different satellite are as follows.

$\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

7 0

4 years ago