which energy does a car travelling 30 m/ph as it slows have:

a). chemical energy
b). thermal energy
c). kinetic energy

please helpp
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Answer:

The additional load is 20g

Answer:

A.

Explanation:

Amplitude measures how much a wave rises or falls. This is illustrated by A.

In the diagram, the amplitude of the wave is shown by A.

There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.

Amplitude is a property that is unique to waves and oscillations.

Therefore, in the diagram, the amplitude of a wave is shown by A.

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2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......

The major principal of leaver is

load × load distance = effort × effort distance

where,

effort dis= distance between effort and fulcrum

load distance = distance between load and fulcrum......

Answer:

freezing point and melting point

Answer:

[tex]\theta=34 \textdegree[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=55kg[/tex]

Angle [tex]\theta =28.0[/tex]

Coefficient of static friction [tex]\alpha =0.680[/tex]

Generally, the equation for Newtons second Law is mathematically given by

For

[tex]\sum_y=0[/tex]

[tex]N=mgcos \theta[/tex]

for

[tex]\sum_x=0[/tex]

[tex]F_{s}=mgsin\theta[/tex]

Where

[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]

[tex]F_{s}=0.68*55*9.8*cos 28[/tex]

[tex]F_{s}=323.62N[/tex]

Therefore

[tex]\alpha mgcos \theta=mg sin \theta[/tex]

[tex]\theta=tan^{-1}(0.68)[/tex]

[tex]\theta=34 \textdegree[/tex]

(a) The static frictional force which holds the box in place is 323.62 N.

(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.

The net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.

∑F = 0

The static frictional force is calculated as follows;

Fs = μFncosθ

Fs = 0.68 x (55 x 9.8) x cos28

Fs = 323.62 N

Fn(sinθ) - μFn(cosθ) = 0

mg(sinθ) - μmg(cosθ) = 0

μmg(cosθ) = mg(sinθ)

μ(cosθ) = (sinθ)

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(μ)

θ = tan⁻¹(0.68)

θ = 34.2⁰

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Answer:

test her prototype and collect data about its flight

Answer:

Mass is zero if the whole rope is horizontal. Imaginary rope.

Explanation:

With any mass at all, only a small section of the rope will be truly horizontal. The rope curve will be a a catenary.

(C)

Explanation:

The capacitance C of a parallel plate capacitor is given by

[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]

Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us

[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

The net force on both crates is the same and the acceleration of both crates is the same.

The given parameters;

The force on the 4kg crate is calculated as follows;

[tex]F_{4kg } = T + F[/tex]

The force on the 6kg crate is calculated as follows;

[tex]F_{6 kg} = -T + F[/tex]

The net force on both crates is calculated as follows;

[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]

Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.

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Answer:

Explanation:

       Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.

           #I AM ILLITERATE

Your Answer Is in this attachment

Answer:

nail file

facial tissues

body wash

deodorant

shampoo

Explanation:

if I'm going by the discount meaning correctly

A theory does not change into a scientific law with the accumulation of new or better evidence. A theory will always remain a theory; a law will always remain a law. Both theories and laws could potentially be falsified by countervailing evidence. Theories and laws are also distinct from hypotheses.

Explanation:

CFL bulbs were made to take the place of incandescent bulbs, which generate light as a result of heat. ... LED (light-emitting diode) is a type of bulb that produces light using a narrow band of wavelengths. LED lighting is more energy efficient than CFL bulb.

Answer:

wavelength= 1.05 × 10^ -46 m

Explanation:

the formula : λ= hc/E

where; "h" = Planck's constant [6.626 × 10^ -34]

c= speed of light [3.0 × 10^ 8]

you first have to convert the energy of the photon to Joules by dividing the constant by 1000

2.09 × 10^ -18 / 1000 = 2.09 × 10^ -21

then you replace you data into the equation

λ= 6.626 × 10^ -34 × 3.0 × 10^ 8 / 2.09 × 10 ^ -21

first multiply the Planck's constant and the speed of light then divide it by the energy which is in "Joules"

:. λ = 1.05 × 10^ -46

hope this helps

b. solve for Vo

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

I hope I helped you ^_^

Answer:

1. F = 569 N

2. F_net = 101N

3. a = 1.74m/s²

Explanation:

Weight is a force measurement.

F = m*a

F = 58.0kg*9.81m/s²

F = 568.98 N

F_net = 670N+(-569N)

F_net = 101N

a = F/m

a = 101N/58.0kg

a = 1.74m/s²

The combined weight of the parachutist and parachute is equal to 568.98 N and the net force due to air resistance exerts a total upward force of 101.02N on her and her parachute, then the magnitude of the acceleration of the parachutist is 1.74 m/s².

Newton's second law states that the resultant force acting on a body is proportional to the rate of change of momentum of that body.

If a  parachutist relies on air resistance to decrease her downward velocity.

The mass of the parachutist and her parachute is 58 kg

The air resistance exerts a total upward force of 670 N

The combined weight of the parachutist and parachute, W = mg

W = 58 × 9.81

W = 568.98 N

The net force on the parachutist = 670 - 568.98 = 101.02 N

The acceleration of the Parachutist = Net force/mass

a = F/m

a = 101.02/58

a  = 1.74 m/s²

Thus, the magnitude of the acceleration of the parachutist would be 1.74m/s².

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Using newtons second law

[tex]\\ \sf\longmapsto Force=Mass\times Acceleration[/tex]

[tex]\\ \sf\longmapsto Force=1(9.8)[/tex]

[tex]\\ \sf\longmapsto Force=9.8N[/tex]

Answer:

It is pure science

Explanation:

A basic knowledge for the discovery of unknown laws based on well controlled experiments and deductions from demonstrated facts or truths.

Answer:

4 device 3 e e and 4 divide 3 + 5 barabar 11 33 size to 46 size 35 and size 49 browser

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

Find the maximum height attained by coin 1 before falling to the ground:

[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]

But the time taken for the coin 1 to reach 11.25 m above the tower:

[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]

The time  at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

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Answer:

Period is 86811.5 seconds.

Explanation:

[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]

[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]

Answer:

the answer is nearly 5.655 [tex]ms^{-2}[/tex]

Explanation:

Given,

[tex]v_{x}=2.7 ms^{-1}[/tex]

[tex]v_{y}=13.3 ms^{-1}[/tex]

[tex]t=2.4 s[/tex]

[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as  [tex]a=\frac{v-u}{t}[/tex])

[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]

[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]

[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]

[tex]=5.655 ms^{-2}[/tex]

hope you have understood this...

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Answer:

Volume of a metal block = 24 cm^3

Volume of a block twice as long, wide and high = 192 cm^3

Explanation:

Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24

Second block, just double each of the lengths to get 6*4*8 = 192

Answer:

SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.

Answer:

variable

Explanation:

Like charges repel and opposite charges attract.