A skater starts from rest v1=0, from the top of an h=6m hill and travels d=20m until it gets to the bottom of the hill where its speed is v2=1.2m/s. If the skater has m=50kg, what is the friction force?

Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific topic.Option A is correct.

A scientific hypothesis is the main element in the scientific method and experiment.

It is said that a hypothesis is a scientific hypothesis. The majority must agree on it, or it must be done after thorough experimentation.

Only scientific inquiries can have their answers provided by scientific endeavor. Scientific concerns have an impact on the research's design as well.

Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific question.

Hence option A is correct.

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Answer:

it alters the rate of chemical reaction

hope it is useful

Answer:

  19.6 m/s

Explanation:

The velocity is the product of acceleration and time.

__

Here, the acceleration is due to gravity.

  v = at

  v = (9.8 m/s²)(2 s) = 19.6 m/s

The stone hits the ground with a speed of 19.6 m/s.

A stone is dropped from the top of a tower which touches the ground after 2 seconds.

Hence, The stone hit the ground with the velocity speed of 19m/s².

- When you're solving velocity, you're determining how fast an object moves from its original position, with respect to a frame of reference, and a function of time.

- That means an object's velocity will be equal to the object's speed and direction of motion.

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The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]

Initial length,[tex]\rm L_i = 7.5 mm[/tex]

[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

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(a) The time taken the block to reach the ground once it flies off the edge of the table is 0.47 s.

(b) The initial horizontal velocity of the block as it flies off the table is 3.91 m/s.

(c) The initial speed of the bullet is 237.5 m/s.

The time of motion of the block is calculated as follows;

h = ¹/₂gt²

where;

t = √(2h/g)

t = √( (2 x 1.08) / 9.8)

t = 0.47 s

x = vt

v = x/t

v = (1.84)/(0.47)

v = 3.91 m/s

m1u1 + m2u2 = v(m1 + m2)

where;

0.00611(u1) + 0.365(0) = 3.91(0.00611 + 0.365)

0.00611u1 = 1.451

u1 = 1.451/0.00611

u1 = 237.5 m/s

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Magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

When the Earth's magnetic field reverses, a new stripe starts to form. Such magnetic patterns enable the scientist to recognize the occurrence of sea-floor spreading and provides strongest evidence for the theory of plate tectonics.

So we can conclude that magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

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Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

The velocity of the third piece is 124.02 m/s at 1.05⁰ below the horizontal.

The velocity of the third piece is calculated from the principle of conservation of linear momentum.

mu = m₁u₁ + m₂u₂ + m₃u₃

where;

6(22)sin(0) = 2(10)sin32 - 1(8)sin(28) + 3u₃y

0 = 6.84 + 3u₃y

u₃y = -6.84/3

u₃y = -2.28 m/s

ΔP = Pf - Pi = J

where;

2640(0.1) = 2(10)cos32 +  1(8)cos(28) + 3u₃x - 6(22)

264 = -108 + 3u₃x

3u₃x = 372

u₃x = 372/3

u₃x = 124 m/s

u₃ = √(124² + 2.28²)

u₃ = 124.02 m/s

tanθ = u₃y/u₃x

tanθ = 2.28/124

tanθ = 0.018

θ = 1.05⁰ (below the horizontal)

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In infrared satellite images, land appears black and clouds appear white. C

Infrared satellite images are images produced as a result of electromagnetic radiations reflected or emitted from a target surface in the infrared position of the electromagnetic spectrum.

On infrared images, the clouds emit very bright white color because of the low temperature and the land appears to be black because of the hot temperature.

Therefore, in infrared satellite images, land appears black and clouds appear white. C

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a. The value of H is 11.51 m

b. The total kinetic energy as she reaches the ground is 6.21 kJ.

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a skier (m=55.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. Value of h = 6.50 m and D = 11.4 m.

Time taken to reach the ground after take off is

t = √2h/g

t = √(2x6.50) /9.81

t =1.15 s

The speed at that time is

speed = distance /time

v = 11.4 /1.15

v = 9.913 m/s

a. From energy conservation principle,

v = √2g(H-h)

Plug the values, we get

9.913 = √(2 x 9.81 (H - 6.5)

Squaring both sides and solving, we have

H = 11.51 m

Thus, the value of H is 11.51 m.

b. The energy at the ground will be only potential energy

Total kinetic energy at the ground = mgH

T.K.E = 55 x 9.81 x 11.51

T.K.E =6.21 kJ

Thus, the total kinetic energy is 6.21 kJ.

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The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.

These key deciding elements are:

1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.

2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.

3. Accumulator identification:

4. Conditions for Installation:

Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

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If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.

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The velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

v = u + at

after 10 seconds;

v = 30 + 10a₁  --(1)

after 5 seconds

v₂ = v + 5a₂ ---- (2)

solve (1) and (2) together

v₂ = 30 + 10a₁ + 5a₂

but a₁/a₂=1.5

a₁ = 1.5a₂

v₂ = 30 + 10(1.5a₂) + 5a₂

v₂ = 30 + 15a₂ + 5a₂

v₂ = 30 + 20a₂

from equation (2), v₂ = v + 5a₂

v + 5a₂ = 30 + 20a₂

v = 15a₂ + 30

Thus, the velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

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Answer:

a community event

Explanation:

The total time the ball is in the air is 1.47 s and the maximum horizontal distance covered is 58.7 m.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

55 = (50sin37)t + ¹/₂(9.8)t²

55 = 30.1t + 4.9t²

4.9t² + 30.1t - 55 = 0

Solve using formula method;

t = 1.47 s

X = vxt

where;

X = (50 x cos37) x 1.47

X = 58.7 m

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Answer: See below

Explanation:

Part (a):

As the velocity of the piano is constant, the net force on the piano is zero. The friction is also zero.

Here, F is the force applied by the man towards the inclined plane, mg is the weight of the piano and N is the normal force.

Applying Newton's law we get,

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

Substituting we get,

[tex]F &= \left( {180\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \sin 19.0^\circ \\ &= 574.3\;{\rm{N}}[/tex]

Therefore, the force is 574.3 N

Part (b)

Here, the force F is applied parallel to the floor.

The friction is zero.

Applying Newton's law we get,

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

Substituting we get,

[tex]F &= \left( {180\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \tan 19.0^\circ \\ &= 607.4\;{\rm{N}}[/tex]

Therefore, the force is 607.4 N

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

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The variable on the y-axis should be the running speed of hamsters since it is the dependent variable.

The dependent variable is the variable whose value fluctuates depending on other variables in an experiment.

In this case, the running speed is the dependent variable because it depends on the weight of the hamster.

The dependent variable is generally graphed on the Y axes, whereas the independent variable is graphed on the X-axes.

In conclusion, the variable on the y-axis should be the running speed of hamsters since it is the dependent variable.

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Answer:

( 688.29 , 276.66 )   meters

Explanation:

Break into horizontal and vertical components

x = 100 cos 55  = 57.36 m/s      

   the x coordinate is then 12 seconds * 57.36 m/s = 688.29 meters

y coordinate is a bit more complex because of the curved path due to gravity

Original y velocity =  100 sin 55

y = vo t - 1/2 (9.81)(t^2)

y =  100 sin 55  (12)  - 1/2 (9.81) (12)^2

      yf = 276.66   meters

Answer:

E = 5.52 x 10⁻¹⁹  J

Explanation:

The formula for energy of a photon, E, is:

[tex]E = hf[/tex]

where h is Plank's constant = 6.62 x 10 ⁻³⁴ .

Using the formula:

E = 6.62 x 10⁻³⁴   x   8.34 x 10¹⁴

  = 5.52 x 10⁻¹⁹  J

Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

Answer:

1.163 seconds

Explanation:

see the attached. solve the quadratic equation.

1. The total mass of the rod is 61.48 kg.

2. The x-coordinate of the center of the mass for this rod is  172.6021 m.

The density is the ratio of the mass and the volume of the object. It is denoted by ρ.

ρ = mass/Volume = m/V

Given a long thin solid rod lies along the positive x-axis. One end is at x = 1.80 m and the other at x = 3.40 m. The linear mass density is

λ = ax³ + bx,

where λ is measured in kg/m, and the constants have the following values: a = 1.70 kg/m⁴ and b = 2.20 kg/m²

So, λ = 1.70x³ + 2.20x

1. The total mass of the rod is

m =₁.₈³⁴ ∫ [1.70x³ + 2.20x] dx

m = [ 1.70x⁴/4 + 2.20x² /2 ]₁.₈³⁴

Putting the limits, we have

m = 69.51 - 8.02548

m = 61.48 kg

Thus, the total mass of the rod is 61.48 kg

2. The x-coordinate of the center of the mass for this rod

Xcm =∫ xλdx/m

Xcm =∫[ x ( 1.70x³ + 2.20x) /m ]dx

Xcm =∫[( 1.70x⁴ + 2.20x²) /m ]dx

On integrating and putting the limits  x = 1.80 m to x = 3.40 m

Xcm = = [ {1.70(3.4)⁵/5 + 2.20(3.4)³ /3} - {1.70(1.8)⁵/5 + 2.20(1.8)³ /3} ]

Xcm = 183.3034 - 10.70133

Xcm = 172.6021 m

Thus,  the x-coordinate of the center of the mass for this rod is  172.6021 m.

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The approximate volume of table tennis ball is  80 cm³

Volume is defined as the amount of space occupied by the three dimensional object. S I unit of volume is m³ or cm³.

To find the volume of tennis ball using graduated cylinder.

Step 1 - Fill the graduated cylinder half or full.

Step 2 - Mark the initial volume of the water i.e. 100 cm³ (Vi)

Step 3  - Put the tennis ball in the graduated cylinder. Some of the water was displaced by the table tennis ball.

Step 4 - Mark the Final volume of the water (Vf) i.e. 180 cm³

Step 5 = Calculate the volume by using Formula

Vb = Vf – Vi = 180 cm³ - 100 cm³ = 80 cm³

Hence the volume of tennis ball (Vb) is 80 cm³

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Answer:

40

Explanation:

The sentence that uses the subjunctive mood correctly is as follows:

Thus, the correct option is D.

Subjunctive mood may be defined as statements or actions that are in the probable situation of doubt and suspicion.

Except for the sentence fourth, the rest all indicates the factual concept with the appropriate condition.

Therefore, the correct option for this question is D.

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Answer:

C. it is important that she apologize to her sister for what she said

Explanation:

test approved

Explanation:

(a) The work done in moving the unit charge from point C to A is 7.62 x 10⁻³ J.

(b) The work done in moving the unit charge from point D to B is 7.62 x 10⁻³ J.

W = Fd

W = Kq²/d

W(C to A) = W(0 to C) + W(0 to A)

[tex]W(C \ to \ A) = - \frac{Kq^2}{7.07} + \frac{Kq^2}{5} \\\\ W(C \ to \ A) = 0.0586 \ Kq^2\\\\W(C \ to \ A) = 0.0586 \times 9 \times 10^9 \times (3.8\times 10^{-6})^2\\\\W(C \ to \ A) = 7.62 \times 10^{-3} \ J[/tex]

W(D to B) = W(0 to D) + W(0 to B)

W(D to B) = 7.62 x 10⁻³ J

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The angle at which it should be banked for a traffic speed of 100 km/h is

14.57°.

To find the angle of banking following formula can be used,

[tex]tan \alpha = v^{2} /rg[/tex]

Given v = 100km/h or

          = 100  x 1000 / 3600

          = 27.7 m/s

Hence,

[tex]tan \alpha = (27.7)^{2} / 300 * 9.81[/tex]

[tex]tan\alpha = 0.26[/tex]

[tex]\alpha = tan^{-1} 0.26[/tex]

[tex]\alpha = 14.57[/tex]°

Hence, the angle of banking of the road will be 14.57°

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The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

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