What is the unique geological feature found on Mercury surface?

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

[tex]T=\dfrac{25}{50}\\\\T=0.5\ s[/tex]

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

[tex]f=\dfrac{50}{25}\\\\f=2\ Hz[/tex]

Hence, this is the required solution.

Answer:

MgS

Explanation:

Answer:

I think it might be (e

Explanation:

OWA OWA

This question is incomplete, the complete question is;

Time Dilation. A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 13.0 months . The scientists need a new laboratory to properly conduct measurements on the isotope. If laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 21.0 months to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00 x 10⁸ m/s.

If they place their sample of the isotope on the spaceship, at what speed must it travel in order for the new laboratory to be completed on Earth by the time half of the isotope on the spaceship decays? Assume that it took one month for the scientists to actually start construction and launch the spaceship so half the sample will remain in 12.0 months but it will still take 21.0 months to build the new lab.

Answer:

the required speed is 2.4618 x 10⁸ m/s

Explanation:

Given the data in the question;

Time taken to complete a lab = 21 months  

it took them one one to actually start so

Time remaining so that sample only decays by 1/2 = (13 - 1) = 12 months

we need to find at what speed the spaceship should travel so that time on earth ( = 12 months) become time on spaceship ( = 21 months)

we make use of time dilation equation;

t' = t/√( 1 - v²/c² )

where t is time in rest frame = 12 and t' is time in moving frame = 21

so we substitute

21 = 12/√( 1 - v²/c² )

21√( 1 - v²/c² ) = 12

√( 1 - v²/c² ) = 12/21

we square both side

( 1 - v²/c² ) = ( 12/21 )²

( 1 - v²/c² ) = 0.3265

v²/c² = 1 - 0.3265

v²/c² = 0.6735

v² = 0.6735 × c²

v = √0.6735 × √c²

v = 0.8206 × c

given that speed of light c = 3 x 10⁸ m/s

v = 0.8206 × 3 x 10⁸ m/s

v = 2.4618 x 10⁸ m/s

Therefore, the required speed is 2.4618 x 10⁸ m/s

work=force×distance this is the formula

Answer:

465 grams of sugar.

Explanation:

I'm not sure this is true: is the relationship linear?

If it is then 300 grams of water should be able told 3*155 grams of sugar

3 * 155 = 465  grams of sugar in 300 grams of water.

Answer:

The formula used to convert from the metric system outside of it. So like converting kilograms into pounds. The Formula is as follows:

# unit x (#unit/#unit) = # unit

   ^                  ^                 ^

   I                   I                  I

given        Conversion  Answer

                     factor

*Note: Italic "units" are the same. Bold "units" are the same.

Example:

One thousand eighty kilometers is how many miles? Set it up dimensionally.

1080 km (1 mi/1.61 km) = 670.81 mi

*This is because 1080 x 1 = 1080, but then you divide 1080 by 1.61

Answer:

27 ms^-1

Explanation:

by using v= u + at

u = 0 ( because the object id starting from rest)

v= 0 + 1.5 x 18

v = 27 ms^-1

Explanation:

A plane mirror always creates an image with the same distance to the mirror as the object, only in the other direction. So both of them have a distance of 10cm, one is 10cm to the left, one 10cm to the right, thus their mutual distance is 20cm

What a fan does is create a wind chill effect. ... By blowing air around, the fan makes it easier for the air to evaporate sweat from your skin, which is how you eliminate body heat. The more evaporation, the cooler you feel.

Answer:

1) the required horizontal force F is 1095.6 N

2) W = 0 J { work done by rope will be 0 since tension perpendicular }

3) work is done by the worker is 1029.4 J

Explanation:

Given that;

mass of bag m = 125 kg

length of rope [tex]l[/tex] = 3.3 m

displacement of bag d = 2.2 m

1) What horizontal force is necessary to hold the bag in the new position?

from the figure below; ( triangle )

SOH CAH TOA

sin = opp / hyp

sin[tex]\theta[/tex] = d / [tex]l[/tex]

sin[tex]\theta[/tex] = 2.2/ 3.3

sin[tex]\theta[/tex] = 0.6666

[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )

[tex]\theta[/tex]  = 41.81°

Now, tension in the string is resolved into components as illustrated in the image below;

Tsin[tex]\theta[/tex] = F  

Tcos[tex]\theta[/tex] = mg

so

Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg

sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg

we know that; tangent = sine/cosine

so

tan[tex]\theta[/tex] = F / mg

F = mg tan[tex]\theta[/tex]

we substitute

Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )

F = 1225 × 0.8944

F = 1095.6 N

Therefore, the required horizontal force F is 1095.6 N

2)  As the bag is moved to this position, how much work is done by the rope?

Tension in the rope and displacement of mass are perpendicular,

so, work done will be;

W = Tdcos90°

W = Td × 0

W = 0 J { work done by rope will be 0 since tension perpendicular }

3) As the bag is moved to this position, how much work is done by the worker

from the diagram in the image below;

SOH CAH TOA

cos = adj / hyp

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]

we substitute

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]  = 1 - h/[tex]l[/tex]

cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]

h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]

h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

now, work done by the worker against gravity will be;

W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

we substitute

W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )

W = 4042.5 × ( 1 - 0.745359 )

W = 4042.5 × 0.254641

W = 1029.4 J

Therefore,  work is done by the worker is 1029.4 J

Answer: Ok so We already know that velocity is on the x-axis.

Since acceleration = Force / Mass

Here the Force is downward due to the gravitational pull or we can say it is along y-axis.

Since acceleration is directly proportional to force, so acceleration is also along y-axis. This means that velocity & acceleration are perpendicular to each other.

Example:

Let us assume that an aeroplane is flying parallel to the horizontal plane. The aeroplane will experience the acceleration in several directions. One of them here is the gravitational pull which is perpendicular to the the apparent velocity. So the net velocity & its direction will depend upon the vector sum total of all the forces/acceleration acting on it. Also because of this gravitational pull the aeroplane rotates along with the earth, which is a proof that the force/g experienced by it does not go waste.

Explanation:

Answer:

A is the answer I think pls check it

Answer:

20mm per second

Explanation:

Answer:

4166.6667m/s

Explanation:

The hour equals 3600 secound, and Km equals to 1000 meter.

Answer:

15000000

Explanation:

formula multiply the lenght value by 1000

Answer:

d. all of the above

Answer:

in my opinion i would say that at home cares of more effective because half of the time your child is safe and sound but in child care programs sometimes you have no idea what is going on when you are not around

Explanation:

Answer:

Horizontal component of pull = (cos 55 x 33.9) = 19.4N.

Net horizontal force = (19.4 - 14.2) = 5.2N.

Work = (fd) = (5.2 x 13.5) = 70.2 Joules.

Rounded to 1 decimal place throughout.

Explanation:

The total work done on the vacuum is 70.2 J.

What is work done?

Work done is equal to product of force applied and distance moved.

Work = Force x Distance

Given is a custodian pulls a vacuum 13.5 m with a 33.9 N force at a 55.0° angle, against a 14.2 N friction force.

Horizontal component of pull = (cos 55 x 33.9) = 19.4N.

Net horizontal force = (19.4 - 14.2) = 5.2N.

Work done by vacuum will be

Work =5.2 x 13.5

Work =70.2 J

Thus, the total work done on the vacuum is 70.2 J.

Learn more about work done

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Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

You don't have a image attached

Answer:

[tex]a=0.13m/s^2[/tex]

Explanation:

From the question we are told that

Mass of first team man [tex]m_1=64kg[/tex]

Force of man first team man [tex]F_1=1350[/tex]

Mass of second team man [tex]m_2=69kg[/tex]

Force of man second team man [tex]F_2=1367N[/tex]

Generally the equation for net force F_n is mathematically given by

[tex]F_n=9(m_1+m_2)a[/tex]

[tex]9(m_1+m_2)a=9(f_2-f_1)[/tex]

[tex]9(64+69)a=9(1367-1350)[/tex]

[tex]a=\frac{9(1367-1350)}{9(64+69)}[/tex]

[tex]a=0.127819m/s^2[/tex]

Therefore the acceleration is given by

[tex]a=0.13m/s^2[/tex]

Answer:

9 electrons

Explanation:

The structure has 9 protons and hence the number of electrons equals the number of protons that's why it is said to be electrical neutral

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

          T² = ([tex]\frac{4\pi ^2}{ G M_s}[/tex] a³ = K_s a³

Let's apply this equation to our case

          a = [tex]\sqrt[3]{ \frac{T^2}{K_s} }[/tex]

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

             10² = K_s  [tex]a_{w}^3[/tex]

             a_w = [tex]\sqrt[3]{ \frac{100}{ K_s} }[/tex]

             a_w = [tex]\frac{1}{ \sqrt[3]{K_s} }[/tex]  4.64

Planet X

             a_x = [tex]\sqrt[3]{ \frac{640^3}{K_s} }[/tex]

             a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

              a_y = [tex]\sqrt[3]{ \frac{80^2}{K_s} }[/tex]

              a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

              a_z = [tex]\sqrt[3]{ \frac{270^2}{K_s} }[/tex]

              a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

Answer:

Look Below -->

Explanation:

a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )

b. Her displacement is  0 km because she went back home. (Displacement is the difference between the end and starting points)

c. 3 km/hr (30 minutes / 10 km)

Answer:

The value 0.51

Explanation:

Answer: Dynamic balance

Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121

(a) The original charge on the 40-pF capacitor is [tex]2 .0 \ \times \ 10^{-8} \ C[/tex].

(b) The charge on each capacitor after the connection is made is [tex]4 .0 \ \times \ 10^{-9} \ C[/tex].

(c)  The potential difference across the plates of each capacitor after the connection is 100 V and 400 V.

The original charge on the 40-pF capacitor is calculated as follows;

[tex]Q = CV\\\\Q = 40 \times 10^{-12} \times 500\\\\Q = 2 .0 \ \times \ 10^{-8} \ C[/tex]

[tex]C = \frac{C_1C_2}{C_1 + C_2} \\\\C = \frac{10 \times 10^{-12} \times 40 \times 10^{-12}}{10\times 10^{-12} \ + \ 40 \times 10^{-12}} \\\\C = 8 \times 10^{-12} \ F[/tex]

[tex]Q = Q_1 = Q_2\\\\Q = 8 \times 10^{-12} \ \times \ 500\\\\Q = 4 \times 10^{-9} \ C[/tex]

The potential difference across the plates of each capacitor after the connection is calculated as follows;

[tex]V = \frac{Q}{C} \\\\V_1 = \frac{Q}{C_1} \\\\V_1 = \frac{4 \times 10^{-9}}{40 \times 10^{-12}} \\\\V_1 = 100 \ V\\\\V_2 = \frac{Q}{C_2} \\\\V_2 = \frac{4 \times 10^{-9}}{10 \times 10^{-12} } \\\\V_2 = 400 \ V[/tex]

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