The weight of a person on Earth is 67.0 N. If the gravitational acceleration on the moon is 1.62 m/s2, calculate the person’s weight on the moon.
Group of answer choices

A) 56.0 N

B) 11.0 N

C) 108 N

D) 4.00 N

Answer:

no sorry

Explanation:

Answer:

yes i play genshin

Explanation:

ignore how this is a year late...what server r u on ?im on america and europe!!!!!!!!!!!!(≧∇≦)

(a) The unknown charge kept at distance 0.250m directly below  -0.450 charge is  2.315 x 10⁻¹² C.

(b) The magnitude of the force is 29.16 x 10¹² N.

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

Given is a negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it .

In equation of the magnitude of force, only the magnitude of charges are taken.

Substituting the values into the expression ,we get the charge

0.150 = (9 x 10⁹ x 0.450x q)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the unknown charge is 2.315 x 10⁻¹² C.

b)Let the unknown charge be  0.450C to exert on -450C.

Substituting the values into the expression ,we get the electrostatic force

F = kq₁q₂/r²

F = (9 x 10⁹ x 0.450x 0.450)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the force is 29.16 x 10¹² N.

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(a) The initial horizontal velocity of the driver is 10 m/s.

(b) The time taken for him to reach the water is 1.75 s.

(c) The horizontal displacement is 17.5 m.

The initial horizontal velocity of the driver is given as 10 m/s.

h = vy(t) + ¹/₂gt²

where;

h = 0 +  ¹/₂gt²

2h = gt²

t = √(2h/g)

t = √(2 x 15 / 9.8)

t = 1.75 s

X = Vx(t)

X = 10 x 1.75

X = 17.5 m

Thus, the initial horizontal velocity of the driver is 10 m/s, the time of motion is 1.75 s and the horizontal displacement is 17.5 m.

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Chains are the type of resistance training equipment which has a fixed range of motion.

This is a type of force which restricts the motion of an object thereby slowering it down.

Chains help to build stamina and power though varying the loads in a fixed range of motion. This is why it was chosen as the most appropriate choice.

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Answer:

c option 22.0A

Explanation:

V=IR

I=R/V

the action-reaction forces that the blocks exert on each other will 13 N and 22N

Force is the product of mass and acceleration. There will be constant acceleration on both object X and Y.

Given that the Mass of block X is 10 kg and that of block Y is 5 kg. The magnitude of force of friction on blocks X and Y is 8.0 N and 4.0 N respectively. if an applied force of 40 N [right] acts on block,

We are given that:

With the use of the formula below,

F - [tex]F_{r}[/tex] = ma

Where

The acceleration will be;

40 - (8 + 4) = (10 + 5)a

40 - 12 = 15 a

28 = 15a

a = 28/15

a = 1.8667 m/[tex]s^{2}[/tex]

The tension between the two blocks will be

Force applied-frictional force on X-tension = mass of X × acceleration

That is,

F - [tex]F_{r}[/tex] = ma

40-8-T = 10 × 1.8667

32 - T = 18.667

T = 32 - 18.667

T = 13.33 N

The opposite reaction will be

F - 8 = 10 x 1.8667

F - 4 = 18.67

F = 18.67 + 4

F = 22.67 N

Therefore, the action-reaction forces that the blocks exert on each other will 13 N and 22N

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Answer:

mass

Explanation:

Mass (M) is the measure of the amount of matter in an object. Mass is measured in grams (g). Mass is measured on a balance by comparing the object against other objects with known masses.

Answer:

Mass is the amount of matter in a substance

The approximate magnitude of the force of air resistance is 882 N

The magnitude of the force of air resistance can be obtained as illustrated below:

F = mg

F = 90 × 9.8

F = 882 N

Thus, approximate magnitude of the force of air resistance is 882 N

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A movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a closed kinetic chain movement.

A closed kinetic chain movement is defined as that position where the most distal aspects of a particular extremity are fixed to the earth or another solid object.

In conclusion; a movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a closed kinetic chain movement.

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The initial speed of the object was 6.26 m/s.

Acceleration of the sled, a = -g*μk

= -9.8*0.2

= -1.96 m/s^2

Let u is the initial speed and v is the final speed.

Applying, v^2 - u^2 = 2*a*d

0^2 - u^2 = 2*(-1.96)*10

u = sqrt(2*1.96*10)

= 6.26 m/s

The kinetic friction coefficient μok is the ratio of the friction force to the normal pressure experienced by a body moving on a dry, non-easy floor. The determination of μok may be achieved quite appropriately through an alternative creative mechanism described with the aid of Timoshenko

The coefficient of friction (fr) is quite a number that is the ratio of the resistive force of friction (Fr) divided with the aid of the everyday or perpendicular force (N) pushing the items collectively. it is represented by the equation: fr = Fr/N.

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Answer:

The acceleration of the rifle equals the force that the bullet exerts on the rifle divided by the mass of the rifle. The two forces are equal, but since the mass of the rifle is much greater than the mass of the bullet, the acceleration of the rifle is much less than the bullet's acceleration.

The magnitude of the force exerted by the second wire is 16.5 N.

Apply cosine rule to determine the force on the second wire;

W² = F₁² + F₂² - 2F₁F₂ cos(θ)

where;

(4.3 x 9.8)² = 31²  +  F₂²  - 2(31) F₂(cos 122)

1775.78 = 961  +  F₂²  + 32.85F₂

814.78 = F₂²  + 32.85F₂

F₂²  + 32.85F₂ - 814.78 = 0

Solve the quadratic equation using formula method;

F₂ = 16.5 N

Thus, the magnitude of the force exerted by the second wire is 16.5 N.

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Answer:

इअग्सिव्ह्ह्व्य्वुव्व्ह्विहेधेइएह

Explanation:

ह्य्स्स्येउ?स्य्ग्स्झ्स

-3.6× 10^5J is the Work done by friction force

Work done = force × distance

v^2 = u^ +2as

u= 22m/s

a =10m/s^2

When the car stops the final velocity (v) =0

0= 22^2 +2×10×s

s = -484/20

s =-24.2m

Work done = force × distance

Force = mass × acceleration

Work done = 1500×10× -24.2

= -3.6×10^5J

Hence the Work done is -3.6×10^5J

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The time period of the pendulum is 2 sec, and the frequency will be 0.5 and will on depend on weight of the object.

The time period of a basic pendulum is defined as the time it takes to complete one full oscillation and is indicated by the letter "T".

This concept of frequency leads to the simplest frequency formula

Frequency = 1 / T.

T signifies the time it takes to complete one wave cycle in seconds, while f denotes frequency.

The formula for time period of simple pendulum is

Time Period = 2π√(l/g)

As we can see there is no mass term in the formula, Hence we can say that the time period of simple pendulum will not depend on mass or weight of the object tied to the string.

We have given length as 1 m

Time Period = 2π√(1/9.8)

Time Period = 2π×0.319

Time Period = 2 seconds

Time of the Simple Pendulum is 2 seconds.

Frequency = 1 / Time period

Frequency = 1 / 2

Frequency = 0.5 1/second

Frequency of the Simple Pendulum is 0.5

So we can conclude that the Time period of Simple pendulum is 2π√(l/g) and as we can see there is no term of mass, means the simple pendulum will not be depend on mass or weight. The time period came out to be 2 seconds and the frequency came out to be 0.5.

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Forward and downward speed are referred to as acceleration and deceleration respectively.

This is the rate at which an object gains speed with respect to time with its unit being meter per second square(m/s²).

Objects in constant speed are said to have a uniform motion. The downward speed is known as deceleration which involves the object losing speed to return to a state of rest.

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1. The part that shows the body is accelerating is the first 2 hours (ABE). The acceleration is 20 Km/h²

2. The part that shows the body is coming to rest the last 2 hours (CDF)

3. The distance travelled in the first 2 hours is 40 Km

Acceleration is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

From the diagram (ABE), we can see that the body starts from rest and attain a final velocity (40 Km/h) within the first 2 hours. This clearly indicates that the body is accelerating.

The acceleration can be obtained as followed:

a = (v – u) / t

a = (40 – 0) / 2

a = 20 Km/h²

From the diagram (CDF), we can see that the object velocity changes from 40 to 0 within the last 2 hours. This clearly indicates that the object caming to rest.

v² = u² + 2as

v² = 2as

s = v² / 2a

s = 40² / (2 × 20)

s = 40 Km

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If they had done a risk assessment first "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it"

An aqueous solution of hydrogen chloride HCl that is a strong corrosive irritating acid, is normally present in dilute form in gastric juice, and is widely used in industry and in the laboratory.

Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.

If they had done a risk assessment first "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it"

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(a) Fundamental frequency of clothesline is 6.98Hz.

(b) Frequency of second harmonic is 13.98Hz.

The wave velocity expression is,

        v = √T / √μ

here, T is tension μ is mass per unit length

     v = √22.1 / √(12.5×10⁻³/7.66)

     v = 116.37 m/s

    For the fundamental frequency,

                  v = fλ

     116.37 m/s= 7.6 ×λ

         λ = 15.31 m

     For the second Harmonic,

          λ₂=  v/ f₂ = 116.37 / 15.2 = 7.66 m

     Now consider mass is increased

             μ= 14.8 × 10⁻³ / 7.66

                v = √T / √μ

                v = √22.1 / √14.80 × 10⁻³/7.66

                v = 106.95 m/s

          Fundamental harmonic,

                f₁ = v/λ₁ = 106.95 / 15.31 = 6.98 Hz

                f₂= v/ λ₂ = 106.95 / 7.65 = 13.98 Hz

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[Your question is incomplete, but most probably your full question was-

a) What is the frequency of the fundamental harmonic for a clothesline with a mass of 14.8 g?

(b) What is the frequency of the second harmonic for a clothesline with a mass of 14.8 g?]

Answer:

kinetic energy and potential and electrical energy

Explanation:

it will be need to maintain the motion of the vehicle and produce electrical energy

The balance point of the bar from the 100 g mass is 12 cm.

Apply principle of moment as follows;

Let the balance point from 100 g mass = x

Clockwise moment = anti clockwise moment

100x = 150(20 - x)

100x = 3,000 - 150x

250x = 3000

x = 3000/250

x = 12 cm

Thus, the balance point of the bar from the 100 g mass is 12 cm.

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The height is 0.70 meters

The formula for gravitational potential energy is

= mgh

m= mass, g= acceleration due to gravity, h= height

54.9= 8 × 9.8 × h

54.9= 78.4h

h= 54.9/78.4

= 0.70 meters

Hence the height at which the bacpack was lifted is 0.70 meters


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Answer:

its 0.7m

Explanation:

Answer:

E = 1/2 * k x^2 is the energy stored in the spring

E = 1/2 * 260 N/m * .08^2 m = .832 Joules

0.29 m/s is the speed of this wave.

How can we determine a wave's speed?

Given: The wave's traveling wavelength is () = 35 cm

The wave's period (t) is 1.2 seconds.

The wave's frequency is 0.833 Hz.

We must ascertain a wave's wave speed.

Formula for wave speed is 35 x 0.833 (wavelength x frequency).

= 29.167 cm/sec

= 0.29 m/sec

As a result, the wave's speed is 0.29 m/s.

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882J of gravitational potential energy is gained by the eagle.

Here, we have been provided with data-

mass, m = 4.5 kg

height, h = 25 m

We have to find the gravitational potential energy.

In the question, it is given that the eagle is at a height of 5m and it flies from there 20 m more to reach its nest.

So, gravitational potential energy is given as

P.E = m x g x h

P.E = 4.5 × 9.8 × 20

P.E =  882J

Thus, the correct option is (b).

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Answer:

b

Explanation:

its b

Answer: Only 25% of the intensity of a polarized light wave passes through a polarizing filter. Then the angle between the electric field and the axis of the filter will be 60 degrees.

Explanation: To find the answer of the problem given, we have to know about the Malu's law of polarization.

                                         [tex]I=I_0cos\alpha[/tex]

                                        [tex]\frac{I}{I_0} =0.25\\[/tex]

                                        [tex]cos^2\alpha =0.25\\cos\alpha =\sqrt{0.25} =0.5\\\alpha =cos^-1(0.5)=60 degrees[/tex]

Thus, we can conclude that, the angle between the electric field and the axis of the filter will be 60 degrees.

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New kinetic energy:

If the initial kinetic energy of a car is 0.3 MJ and the speed of the car decreased two times then the new kinetic energy will be 0.075 MJ.

The calculation for kinetic energy:

Step-1:

It is given that the car has a kinetic energy of 0.3 MJ. It is required to find the new kinetic energy when the car's speed decreases two times from its initial speed.

Let the mass of the car is m and the initial speed of the car is v.

It is known that the kinetic energy is calculated as,

[tex]\text{K.E}=\frac{1}{2}\times \text{mass}\times\text{speed}^2[/tex]

Therefore the initial kinetic energy is the car is,

[tex]\text{K.E}_1=\frac{1}{2}\times \text{m}\times\text{v}^2[/tex]

Step-2:

The speed of the car is decreased two times. Thus the speed becomes v/2. Therefore the final kinetic energy is the car is,

[tex]\text{K.E}_2&=\frac{1}{2}\times \text{m}\times(\frac{v}{2})^2\\&=\frac{1}{4}\times\frac{1}{2}\times \text{m}\times\text{v}^2\\&=\frac{1}{4}\times\text{K.E}_1\\&=\frac{1}{4}\times {0.3}\text{ MJ}\\&=0.075 \text{MJ}[/tex]

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Answer:

yes

Explanation:

banked curve :

is a road that looks like the top part of a race track

its looks elevated

often seen in bicycle race track (velodrome)

Banking the curve can help keep cars from skidding. When the curve is banked, the centripetal force can be supplied by the horizontal component of the normal force.

sfuca