what happens to the weight of the body when it is falling freely under the action of gravity​

Answers

Answer 1

Answer:

A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object

Explanation:

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Answer 2

Answer:

Gradually increases until the maximum weight reaches the surface of the earth

Explanation:


Related Questions

A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.

Answers

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)

Answers

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

If "FRICTION" means breaking up of relationships, then how can you reduce friction among friends?

A. Respect the opinion of your friends
B. Learn how to be understanding
C. Avoid bullying your friends
D. Force your friends to do things they don't like​

Answers

Answer:

A

Explanation:

respect the opinion of your friends

a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car

Answers

Answer:

a) T = 0.5 s

b) v = 1.2π m/s ≈ 3.77 m/s

Explanation:

It makes two revolutions in one second so makes one revolution in ½ second

circumference of the circle is

C = 2πr = 0.6π m

which it traverses in one time period

0.6π m / 0.5 s = 1.2π m/s

To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

What is speed?

Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.

Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.

Mathematically, the formula for speed can be given as

speed= distance/time

Values that are given

Time period= 0.5 s

Circumference = 2πr = 0.6π m

substituting all the given values in the above equation, we get

speed     =0.6π m / 0.5 s

On calculations, we get

              = 1.2π m/s

              =3.77 m/s

Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

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A brass road is 2cm long at instance to what is the lense for a temperature rise of 100k, If the expansivity of brass is 18x10^-6/k^-1

Answers

The length of the brass at a temperature rise of 100 K is 2.0036 m

From the question given above, the following data were obtained:

Original length (L₁) = 2 m

Temperature rise (ΔT) = 100 K

Coefficient of linear expansion (α) = 18×10¯⁶ K¯¹

Final length (L₂) =?

The final length of the brass can be obtained as follow:

α = L₂ – L₁ / L₁ΔT

18×10¯⁶ = L₂ – 2 / (2 × 100)

18×10¯⁶ = L₂ – 2 / 200

Cross multiply

L₂ – 2 = 18×10¯⁶ × 200

L₂ – 2 = 0.0036

Collect like terms

L₂ = 0.0036 + 2

L₂ = 2.0036 m

Thus, the length of the brass at a temperature rise of 100 K is 2.0036 m

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Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?

Answers

Answer:

[tex]_{18}^{39} } Ar[/tex]

Explanation:

The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.

[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]

Argon is a stable gas and is found in the group 8 on the periodic table of elements.

Answer:

Answer is below

Explanation:

39 18 Ar

The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.

Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?

Answers

Answer:

a) required area is 1.1318 m²

b) the maximum potential difference that can be applied across the compactor is 1931.1 V

Explanation:

Given the data in the question;

dielectric constant εr = 2.35

distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m

dielectric strength = 49.5 MV/m

a)

given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F

To find the Area, we use the following the expression.

C = ε₀εrA / d

we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹²  (F/m)

we substitute

0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A  ] /  7.85 × 10⁻⁵

A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]

A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹

A = 1.1318 m²

Therefore, required area is 1.1318 m²

b)

the maximum potential difference that can be applied across the compactor.

We use the following expression;

⇒ 1/2 × dielectric strength × thickness d

we substitute

⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )

1931.1 V

Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V

One ball is aprojected in the uapward directon with a certain velocity ‘v’ and other

is thrown downwards with the same velocity ​

Answers

Complete question is;

One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.

The ratio of their potential energies at highest points of their journey, will be:

Answer:

u² : (u cos θ)²

Explanation:

Maximum potential energy for the first ball will be at a maximum height of;

H = u²/2g

Thus;

PE = mg(u²/2g)

For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g

PE = mg((u cos θ)²/2g)

The ratios of the potential energies are;

mg(u²/2g) : mg((u cos θ)²/2g)

mg will will cancel out since they are of same mass.

Thus;

(u²/2g) : (u cos θ)²/2g

Again 2g will cancel out to give;

u² : (u cos θ)²

I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)


Solve for a

Answers

Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

What is the feature known as the "Great Dark Spot" of Neptune? It is an apparently permanent feature about five times the size of Earth, similar to the Great Red Spot of Jupiter, near Neptune's south pole. It was a dark hole in the upper atmosphere left by the collision of the comet Shoemaker-Levy 9. It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years. It is a dark surface feature on the surface snow layers caused by radiation discoloration of the older layers. It is a permanent discoloration of the north polar region of Neptune caused by locally prevailing lower surface temperatures there.

Answers

Answer:

It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.

Explanation:

The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.

The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.

A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of the magnetic field if there is a 45o angle between it and the proton's velocity

Answers

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

[tex]F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T[/tex]

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?

Answers

Answer:

The wavelength stays the same.

Explanation:

When the amplitude is increased, the wavelength stays the same.

Here the wavelength doesn't depend upon the amplitude.

Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one

Answers

Answer:

F' = 16 F

Hence, the electric force between charges becomes sixteen times its initial value.

Explanation:

The electric force between the two charges is given by the Colomb's Law:

[tex]F = \frac{KQ_1Q_2}{R^2}[/tex] ------------------- eq(1)

where

F = electric force

K = Colomb's Constant

Q₁ = magnitude of the first charge

Q₂ = magnitude of the second charge

R = Distance between charges

Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:

[tex]F' = \frac{K(2Q_!)(2Q_2)}{(\frac{R}{2})^2}\\\\F' = 16 \frac{KQ_1Q_2}{R^2}[/tex]

using equation (1):

F' = 16 F

Hence, the electric force between charges becomes sixteen times of its initial value.

If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25

Answers

Answer:

100Hz

Explanation:

In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz

Answer:

HZ 100 is the right answer hope you like it

Outline five ways of varying the force on a current-carrying conductor in a magnetic field. (7 marks) ​

Answers

Is this it? I think it is :)

write state of matter with 5 example of each

Answers

There are broadly 3 states of matter (there are 5, but they don't teach 2 of them at school).

1. Solid

Examples: Iron, wood, steel, ice, paper

2. Liquid

Examples: Water, mercury, milk, soup, juice

3. Gas

Examples: Oxygen, Chlorine, Carbon dioxide, Sulphur dioxide, Nitrogen

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine

Answers

Answer:

The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".

Explanation:

As we know,

PV = nRT

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]

then,

⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]

⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]

        [tex]=\frac{10}{3} \ Bar[/tex]

Thus the above is the correct answer.

Calculate area moment of inertia for a circular cross-section with 3 mm diameter:

Answers

Answer:

circles

A=7.07multiple 10-6 m2

I hope you understand and help

The area of the circular cross-section will be 7068 × [tex]10^{-6}m^{2}[/tex].

What is the area?

The measurement that represents the size of a region on a plane or curved surface is called an area.

What is cross-section?

A cross-section would be the non-empty point where a solid body intersects a plane in three dimensions or its equivalent in higher dimensions.

Given data:

Diameter = 3 × [tex]10^{-3} m[/tex].

It is known that. Diameter = 2 radius.

The area can be calculated by using the formula:

A = 1/4 [tex]\pi[/tex][tex]d^{2}[/tex] = 1/4 (3.14) [tex](3 * 10^{-3})^{2}[/tex]= 7068 × [tex]10^{-6}m^{2}[/tex].

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A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?

Answers

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds


(I was only able to do A and B)

Answer:

Explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

which option is correct n why?
6. The projectile motion is a good example of
A. one dimensional motion.
B. two dimensional motion.
C. three dimensional motion.
D. four dimensional motion.

Answers

2. two dimensional motion

Because it has just 2 dimensions x and y

Two dimensional motion

Urgent please help me

Answers

1433 km

Explanation:

Let g' = the gravitational field strength at an altitude h

[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]

We also know that g at the earth's surface is

[tex]g = G\dfrac{M_E}{R_E^2}[/tex]

Since g' = (2/3)g, we can write

[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]

Simplifying the above expression by cancelling out common factors, we get

[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]

Taking the square root of both sides, this becomes

[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]

Solving for h, we get

[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]

[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]

distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh​

Answers

The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.

Answer:

25 you said ? thats incorecct

Explanation:

How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V

Answers

a) A charged particle accelerates as it moves from location A to location B. If V A = 170 V and V B = 210 V, what is the sign of the charged particle? positive negative (b) A proton gains electric potential energy as it moves from point 1 to point 2.
Physics
Two identical point charges are fixed to diagonally opposite corners of a square that is 1.5 m on a side. Each charge is +2.0 µC. How much work is done by the electric force as one of the charges moves to an empty corner? I have W(fe)= -EPE=-q[V(f)-V(i)].
physics
Two point charges are placed on the x axis. (Figure 1) The first charge, q1 = 8.00nC , is placed a distance 16.0m from the origin along the positive x axis; the second charge, q2 = 6.00nC , is placed a distance 9.00m from the origin along the negative x

Infrared and ultraviolet waves have different frequencies.
Both types of wave can have harmful effects on human beings.

Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves.​​

Answers

Answer:

For infrared and ultraviolet waves have different frequencies. Both types of wave can have harmful effects on human beings. Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves. Medical studies indicate that prolonged IR exposure can lead to lens, cornea and retina damage, including cataracts, corneal ulcers and retinal burns, respectively. To help protect against long-term IR exposure, workers can wear products with IR filters or reflective coatings.When you look at the EM spectrum, UV waves are quite a bit smaller in wavelength than infrared, and x-rays/gamma rays are even smaller. Therefore, UV waves are probably causing more harm than infrared waves, and x-rays/gamma rays are probably doing even more damage.

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Infrared and ultraviolet waves have different frequencies. Infrared waves have lower frequencies and longer wavelengths, while ultraviolet waves have higher frequencies and shorter wavelengths.

Harmful effects of Infrared waves:

Infrared waves have lower frequencies and are often associated with heat radiation. Prolonged exposure to intense infrared radiation can lead to thermal burns and damage to the skin and eyes. Infrared radiation can also cause dehydration and overheating of the body, especially in hot environments. While infrared radiation is not as harmful as ultraviolet radiation, excessive exposure can still lead to health issues.

Harmful effects of Ultraviolet waves:

Ultraviolet waves have higher frequencies and shorter wavelengths, making them more energetic than infrared waves. UV radiation from the sun is a well-known harmful agent. Short-term exposure to intense UV radiation can cause sunburn, skin redness, and eye irritation. Long-term exposure to UV radiation can lead to more serious health problems such as skin aging, cataracts, and an increased risk of skin cancer. UV radiation can also damage DNA in skin cells, leading to mutations and potential carcinogenesis.

It is essential to protect ourselves from both infrared and ultraviolet waves to prevent harmful effects. Using sunscreen and wearing protective clothing can help shield the skin from UV radiation. Limiting exposure to intense sources of infrared radiation, such as hot objects or infrared heaters, can help reduce the risk of thermal burns and overheating. Understanding the differences in the frequencies of these waves allows us to implement appropriate safety measures and protect our health.

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I NEED YOUR HELP!!
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.


52 cm4


72 cm4


32 cm4


24 cm4


2 cm4

Answers

Answer:

Minimum area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum area of rectangle

Computation :

Minimum area of rectangle = Length of rectangle x Width of rectangle

Minimum area of rectangle = 6 x 4

Minimum area of rectangle = 24 centimeter²

A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
a. What is the effective spring constant for this motion?
b. How much energy is involved in this motion?

Answers

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?​

Answers

Answer:

Explanation:

Kinematic equation

v = u + at

If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.

a) v = 6  + (-10)(0.5) = 1 m/s    the result is positive, so upward

b) v = 6  + (-10)(2) = -14 m/s    the result is negative, so downward

1. A flywheel begins rotating from rest, with an angular acceleration of 0.40 rad/s. a) What will its angular velocity be 3 seconds later? b) What angle will it have turned through in that time? ​

Answers

Answer:

(a) 1.2 rad/s

(b) 1.8 rad

Explanation:

Applying,

(a) α = (ω-ω')/t................ Equation 1

Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.

From the question,

Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)

Substitute these values into equation 1

0.40 = (ω-0)/3

ω = 0.4×3

ω = 1.2 rad/s

(b) Using,

∅ = ω't+αt²/2.................. Equation 2

Where ∅ = angle turned.

Substitutting the values above into equation 2

∅ = (0×3)+(0.4×3²)/2

∅ = 1.8 rad.

write a use of magnetic force and frictional force each​

Answers

Magnetic force:Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. ... Frictional force:Quite often uses of friction can be seen from how things would be without friction.

The speed of a car decreases uniformly as it passes a curve point where normal component of acceleration is 4 ft/sec2. If the car total acceleration of 5ft/sec2 is the same as it passes a hump, the tangential component of acceleration is _______________ ft/sec2.

Answers

Answer:

45

Explanation:

ft/sec2

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