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2 years ago

10

If a planet's orbital speed is 20 km/s when it's at its average distance from the sun which is most likely orbital speed when it

is nearest the sun?

1 answer:

Alona [7]2 years ago

8 0

Answer:25km/s

Explanation:

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A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

3 0

3 years ago

Read 2 more answers

How can liquid pressure phenomena be used in daily life. Give one example with explanation of working?

N76 [4]

Answer:

An example in which liquid pressure phenomena can be used in daily life is in Water blasting

Explanation:

Water blasting refers application of pressurized water to remove materials from the surface of objects.

There are different varieties of water blasting, including;

Hydrocleaning; Cleaning enabled by the use of high pressure water

Hydrodemolition; Demolition or removal of concrete using pressurized water

Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.

8 0

3 years ago

Sharks and related fish can sense the extremely weak electric fields emitted by their prey in the surrounding waters. These dete

Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

$\Delta V=Ed$ (1)

where

$\Delta V$ is the potential difference

E is the electric field

d is the distance

We want to know the distance the detectors have to be placed in order to achieve an electric field of

$E=1 V/cm=100 V/m$

when connected to a battery with potential difference

$\Delta V=1.5 V$

Solving the equation (1) for d, we find

$d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m$

5 0

3 years ago

A force F~ = Fx ˆı + Fy ˆ acts on a particle that

Hatshy [7]

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

$tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\$

Then we find the angle β of the displacement vector

$tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\$

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

$W=F*d*cos (theta)$

The absolute value of F will be:

$F=\sqrt{8^{2}+1^{2} } \\F= 8.06 N$

The absolute value of d will be:

$d=\sqrt{(6 )^{2}+(2)^{2} } \\d= 6.32m\\$

Now we have:

$W=8.06*6.32*cos(25.56)\\W=46 J$

4 0

3 years ago

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

So

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is $\sqrt{3^2+(-5)^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$

3 0

3 years ago