All categories

3 years ago

Give two example for push or pull to change the state of motion of time two examples

Physics

1 answer:

dolphi86 [110]3 years ago

4 0

Explanation:

An Example of push as a force would be to push on a swing. The force moves the swing in a particular direction and the harder that you push the further the swing will go.

An example of pull as a force would be opening a door. ...

An example of pressure as a force is when you push down on a pile of grapes. is this what you mean

You might be interested in

Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe

Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

$T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$ where is k is the proportionality constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence $T_{1} =365.5 days$

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence $R_{1} =1 AU$

The distance of venus from sun is 0.723 AU i.e $R_{2} =0.723$

From keplers law we know that- $\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }$

⇒ $T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }$

Putting the values mentioned above we get-

$T_{2} ^{2} =50,350.132851075$

⇒ $T_{2} =\sqrt{50,350.132851075}$

⇒ $T_{2} = 224.388352752710 days.$

Hence the time period of venus is 224.388352752710 days

7 0

4 years ago

Read 2 more answers

In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used. The two thermocouples in each sample are

sergeinik [125]

Answer:

K = .3941 × 10³ W/m.K

Explanation:

Qcond = K A ΔT÷ L

∴K = Qcond ×L ÷ A ΔT

J ÷ S = P

P = I × V =Qcond

∴Qcond = I × V

= 0.6 A × 110 V

=66 W

L = 0.12 m

ΔT = 8 °C

Qcond =33 V

Area = (πD²) ÷ 4

= [π (4 × 10⁻² )²] ÷ 4

= 1.256 × 10⁻³ m²

∴A = 1.256 × 10 ⁻³³ m²

So K = ( Qcond × L ) ÷ A ΔT

= (33) (0.12 ) ÷ (1.256 ×10⁻³ ) × 8

= 0.3941 × 10³ W/m .K

7 0

3 years ago

What quantities does angular momentum depend upon?

jarptica [38.1K]

Mass, velocity, and radius that the answer

5 0

3 years ago

When opening a door, you push on it perpendicularly with a force of 57.0 n at a distance of 0.470 m from the hinges. what torque

TiliK225 [7]

<span>Torque = force * distance from axis of rotation. Multiply. If it's not perpendicular, you multiply that with the sine of the angle between the force and the perpendicular direction. T= 0.470*57*Sin 90 = 26.79 N</span>

3 0

3 years ago

The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is

yKpoI14uk [10]

We use the voltage division problem between the load resistance, amplifier output resistance as

$V_{out} = V_{amlifire} \times \frac{R_{load} }{R_{load} + R_{out} }$ .

Here, $V_{out}$ is the output voltage, $V_{amlifire}$ is the amplifier voltage, $R_{load}$ is the load resistance and $R_{out}$ is the amplifier output resistance.

Therefore,

$1-\frac{20}{100} = \frac{1 \ k\Omega }{1 \ k\Omega +R_{out} } \\\\ R_{out} = \frac{1 \ k\Omega }{0.8} -1 \ k\Omega =1250 \Omega -1000 \Omega =250 \Omega$ .

Thus, the amplifier output resistance is $250 \ \Omega$ .

4 0

3 years ago

Give two example for push or pull to change the state of motion of time two examples​

Give two example for push or pull to change the state of motion of time two examples