a hiker walks with an average speed of 2.2 m/s what distance in kilometers does the hiker travel in a time of 2.4 hours.​

Answer:

Kinetic energy

Explanation:

Kinetic energy because its energy of motion and when we usually clap our hands it makes noise

Answer:

I am going to guess it shows that the balloon is going downwards because the speed of rise is in the negatives for the last 2.

Answer:

If the eclipse is a total lunar eclipse, the Moon will pass through the umbra (area of total shadow) created by Earth over the course of about two hours. ... A partial eclipse of the Moon occurs when the Moon passes through only part of Earth's umbra or only its penumbra.

Answer:

Not one of your choices .....closest is 90 N

Explanation:

Saturn g = 10.44 m/s^2

10 kg * 10.44 m/s^2 = 104.4 N

What is the angle of reflection? 45 degrees

Answer: 45 degrees

Answer:

Sodium fluoride

Explanation:

Sodium Fluoride

Applying ohm's law

[tex]\\ \sf\longmapsto \dfrac{V}{I}=R[/tex]

[tex]\\ \sf\longmapsto I=\dfrac{V}{R}[/tex]

[tex]\\ \sf\longmapsto I=\dfrac{5}{10}[/tex]

[tex]\\ \sf\longmapsto I=0.5A[/tex]

Answer:

from

velocity= displacement/time

then

time=displacement/velocity

time= 4200/70

time=60s

time will take 60seconds

Table tennis can be defined as an indoor sport and recreational activity in which two (2) or four (4) players hit a ping-pong ball back and forth on a table that is divided into halves by a low net, especially through the use of a small-solid bat (racket).

Generally, there are two (2) main types of grip in table tennis and these include:

Basically, there are four (4) fundamental skills used in table tennis and these are:

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The tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

The Principle of Moments states that when a body is in equip, the sum of clockwise moment about a point is equal to the sum of anticlockwise moment about the same point.

The formula for calculating moment is given below:

From the principle of moments:

Let tension in chain 1 be T1 and tension in chain 2 be T2.

T1 + T2 = 150 + 650 + 419

T1 + T2 =1219

Taking all distances from chain 1,

Sum of Moments = 0

419 × 0.5 + 150 × 0.85 + 650 × 0.9 = T2 × 1.7

T2 = 922/17

T2 = 542.35 N

Then, T1 = 1219 - 542.35

T1 = 676.65 N

Therefore, the tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

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Using Newton's second law

[tex]\\ \tt\longmapsto F=ma[/tex]

[tex]\\ \tt\longmapsto 20000=23m[/tex]

[tex]\\ \tt\longmapsto m=20000/23[/tex]

[tex]\\ \tt\longmapsto m=869.5kg[/tex]

This question involves the concepts of density, weight, and volume.

The change in weight of the cylinder will be "2.72 x 10⁻³ N".

The change in weight of the cylinder will be equal to the weight of the air inside the cylinder which is being removed from it:

[tex]\Delta W = W_a[/tex]

where,

Therefore,

[tex]\Delta W = W_a = \rho Vg\\\\\Delta W = (1.225\ kg/m^3)(2.26\ x\ 10^{-4}\ m^3)(9.81\ m/s^2)[/tex]

ΔW = 2.72 x 10⁻³ N

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Answer:

8s

Explanation:

period of pendulum = 2π*√(l/g)

l = length of pendulum

period for 1st pendulum = 4 sec

thats 2π*√(l/g) = 4

for 2nd pendulum

length = 4 l

substitute 4l in formula

2π*√4l/g = 2*4= 8s

Given the data from the question, the period of the pendulum 2 is 8 s

The period of pendulum 2 can be obtained as illustrated below:

T²₁ / L₁ = T²₂ / L₂

4² / L = T²₂ / 4L

Cancel out L

4² = T²₂ / 4

Cross multiply

T²₂ = 4² × 4

T²₂ = 64

Take the square root of both sides

T₂ = √64

T₂ = 8 s

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Answer:

Momentum  P is 840000kgm/s or 8.4 × 10^6

Explanation:

Data :

        Mass = 21000 kg

        Velocity = 400 m/s

So momentum is given as

      P = mv

      P = 21000×400

    P = 8400000 kgm/s

       P = 8.4 × 10^6

If a car traveling with an initial velocity of 31 meters/seconds slows down at a constant rate of 5.4 meters/seconds for 3 seconds, the velocity at the end would be 47.2 m/s.

There are three equations of motion given by  Newton,

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

As given in the problem , a car traveling with an initial velocity of 31 meters/seconds slows down at a constant rate of 5.4 meters / seconds² for 3 seconds,

By using the second equation of motion,

S = ut + 1/2*a*t²

  =31×3 + 0.5×5.4×3²

  = 93 + 24.3

  =117.3

Now by using the third equation of motion,

v² - 31² = 2×5.4×117.3

v = 47.2 m/s

Thus, the velocity of the car at the end of the three seconds would be 47.2 m/s.

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With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x [tex]10^{11}[/tex] N and 66 degrees

ELECTRIC FORCE (F)

F = [tex]\frac{KQq}{d^{2} }[/tex]

Where K = 9 x [tex]10^{9}[/tex] N[tex]m^{2}[/tex]/[tex]C^{2}[/tex]

The distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex] can be calculated by using Pythagoras theorem.

d = [tex]\sqrt{33^{2} + 33^{2} }[/tex]

d = 46.7 cm = 0.467 m

For force [tex]F_{1}[/tex], substitute all the parameters into the formula above

[tex]F_{1}[/tex] = (9 x [tex]10^{9}[/tex] x 3 x 1)/[tex]0.467^{2}[/tex]

[tex]F_{1}[/tex] = 2.7 x [tex]10^{10}[/tex]/0.218

[tex]F_{1}[/tex] = 1.24 x [tex]10^{11}[/tex] N

For force [tex]F_{4}[/tex], substitute all the parameters into the formula above

[tex]F_{4}[/tex] = (9 x [tex]10^{9}[/tex] x 3 x 4)/[tex]0.33^{2}[/tex]

[tex]F_{4}[/tex] = 1.08 x [tex]10^{11}[/tex]/0.1089

[tex]F_{4}[/tex] = 9.92 x [tex]10^{11}[/tex] N

For force [tex]F_{2}[/tex], substitute all the parameters into the formula above

[tex]F_{2}[/tex] = (9 x [tex]10^{9}[/tex] x 3 x 2)/[tex]0.33^{2}[/tex]

[tex]F_{2}[/tex] = 5.4 x [tex]10^{10}[/tex]/0.1089

[tex]F_{2}[/tex] = 4.96 x [tex]10^{11}[/tex] N

Summation of forces on Y component will be

[tex]F_{y}[/tex] = [tex]F_{4}[/tex] - [tex]F_{1}[/tex] Sin 45

[tex]F_{y}[/tex] = 9.92 x [tex]10^{11}[/tex] - 1.24 x [tex]10^{11}[/tex] Sin 45

[tex]F_{y}[/tex] = 9.04 x [tex]10^{11}[/tex] N

Summation of forces on X component will be

[tex]F_{x}[/tex] = [tex]F_{2}[/tex] - [tex]F_{1}[/tex] Cos 45

[tex]F_{x}[/tex] = 4.96 x [tex]10^{11}[/tex] - 1.24 x [tex]10^{11}[/tex] Sin 45

[tex]F_{x}[/tex] = 4.08 x [tex]10^{11}[/tex] N

Net Force = [tex]\sqrt{F_{x} ^{2} + F_{y} ^{2} } }[/tex]

Net force = [tex]\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }[/tex]

Net force = 9.9 x [tex]10^{11}[/tex] N

The direction will be

Tan ∅ = [tex]F_{y}[/tex]/[tex]F_{x}[/tex]

Tan ∅ = 9.04 x [tex]10^{11}[/tex] / 4.08 x [tex]10^{11}[/tex]

Tan ∅ = 2.216

∅ = [tex]Tan^{-1}[/tex](2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x [tex]10^{11}[/tex] N and 66 degrees approximately.

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The addition of vectors and Coulomb's law allows us to find the result for the force on the charge q₃ is:

Given parameters,

To find.

The electric force is given by Coulomb's law, which states that the force is proportional to the charges and inversely proportional to the square of the distance.

               F =[tex]k \frac{q_1q_2}{r_{12}^2}[/tex]  

where F is the force, k Coulomb's constant, q the charges and r the distance between the charges.

Force is a vector magnitude, so vector algebra must be used for vector addition, an easy and efficient way is to use an analytical method for addition:

In the attachment we have a diagram of the forces, the sum on each axis is:

x axis

          Fₓ= F₂- F₁ₓ

y axis

          [tex]F_y = -F_4 + F_{1y}[/tex]  

Let's use trigonometry to find the component of force.

        cos 45 = [tex]\frac{F_1_x}{F_1}[/tex]  

        cis 45 = [tex]\frac{F_1_y}{F_1}[/tex]  

        F₁ₓ= F₁ cos 45

        [tex]F_1_y[/tex] = F₁ sin 45

Let's look for the distance between the charges, for charges 2 and 3 charges 4 and 3 the distance of is equal to the side of the square

          r=r₂₃ = r₄₃ = d

The distance between charges 1 and 3 is the diagonal of the square that we can find with the Pythagorean theorem.

       R₁₃² = d² + d²

        R₁₃² = 2d²

We substitute in Coulomb's law.

        F₂₃ = [tex]k \frac{q_2q_3}{d^2}[/tex]

        F₄₃ = [tex]k \frac{q_4q_3}{d^2}[/tex]

         F₁₃ = [tex]k \frac{q_1q_3}{2d^2 }[/tex]  

Let us substitute in the components of the force on each axis.

x axis

         Fₓ= F₂- F₁ₓ

         [tex]F_x = k \frac{q_2q_3}{d^2} - k \frac{q_1q_3}{2d^2} cos 45 \\F_x = k \frac{q_3}{d^2} ( q_2 - \frac{q_1 cos 45}{2})[/tex]

y axis

          [tex]F_y = - F_4 + F_1_y \\F_y = k \frac{q_4q_3}{d^2} - k \frac{q_1q_3}{2d^2} sin 45 \\F_y = k \frac{q_3}{d^2} ( -q_4 + \frac{q_1sin 45}{2})[/tex]

We substitute in value of each charge.

x axis

           [tex]F_x = k \frac{3q}{d} \ q( 2 - \frac{1 \ cos 45}{2} ) \\F_x = 3k (\frac{q}{d})^2 1.6464[/tex]

y axis

         [tex]F_y = k \frac{3q}{d^2} ( -4 + \frac{1 \ sin45}{2} ) \\F_y = 3k (\frac{q}{d})^2 ( -3.64645)[/tex]

The resulting vector is

          F = [tex]F_x \hat i + f_y \hat j[/tex]  

          F = [tex]3k(\frac{q}{d} )^2 ( 1.6464 \hat i - 3.64645 \hat j )[/tex]3k q² /d² ( 1.6464 i^ - 3.64645 j^ )

We calculate.

        F = 3 9 10⁹ ( 2.4 10-6/0.33)² ( 1.6464 i^ - 3.64645 j^ )

         F = 1.428( 1.6464 i⁻ 3.64645 j^

We build the resulting vector, for the module we use the Pythagorean theorem.

          [tex]F = \sqrt{F_x^2 + F_y^2}\\F = 1.428 \ \sqrt{1.6464^2+ 3.64645^2}[/tex]

          F = 5.71N

Let's use trigonometry for the angle.

        [tex]tan \theta = \frac{F_y}{F_x}[/tex]  

        [tex]\theta = tan^{-1} (\frac{-3.64645}{1.6464} )[/tex]  

        θ = -65º

This angle is half clockwise from the positive side of the x-axis.

in conclusion using vector addition and Coulomb's law we can find the result for the force on the charge q3 is:

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Answer:

According to ohms law

V= I x R

R= V ÷ I

R= 110 ÷ 2

R= 55 ohms

Explanation:

Answer:

1/f = 1/o + 1 /i       thin lens equation

1 / i = 1/f - 1/o = (o - f) / (o f)    rearranging

or i = o f / (o - f)   for the image distance

This image will be positive (real) with values given

m = -i / o      magnification

m = - f / (o - f)      shows m is negative and the image inverted

Also, since o here is large m will be small and the image is small

I think they mean A as the right answer

Different texts use different terms but here i is image distance and o the object distance

Answer: D) Big - straight

(a) The speed of the mass m2 just before it hits the floor for frictionless table is 1.98 m/s.

(b) The speed of the mass m2 just before it hits the floor when there is friction is 1.53 m/s.

[tex]-f_k + T = m_1a\\\\ T = m_1 a + f_k \ ---\ (1)\\\\ [/tex]

[tex]m_2g - T = m_2a \ ---- (2)\\\\ [/tex]

Solve (1) and (2) together

[tex]m_2 g - (m_1 a+ f_k) = m_2 a\\\\ m_2 g - f_k = m_1 a + m_2 a\\\\ m_2 g - f_k = a (m_1 + m_2)\\\\ a = \frac{m_2 g- f_k}{m_1 + m_2} [/tex]

When the table is frictionless, the acceleration of the masses is calculated as follows;

[tex]a = \frac{m_2 g}{m_1 + m_2} \\\\ a = \frac{1 \times 9.8}{2 + 1} \\\\ a = 3.27 \ m/s^2[/tex]

When the coefficient of friction between m1 and the table is 0.2, the acceleration of the masses is calculated as follows;

[tex]a = \frac{m_2 g - f_k}{m_ 1+ m_2} \\\\ a = \frac{m_2 g - \mu_k m_1g}{m_1 + m_2} \\\\ a = \frac{g(m_2 - \mu_km_1)}{m_1 + m_2} \\\\ a = \frac{9.8(1 - \ 0.2\times 2)}{2+1 } \\\\ a = 1.96 \ m/s^2[/tex]

The height of the mass m2 above the ground = 60 cm = 0.6

The speed of the mass m2 just before it hits the floor for frictionless table is calculated as follows;

[tex]v^2 = u^2 + 2ah\\\\ v^2 = 0 + 2ah\\\\ v^2 = 2ah\\\\ v= \sqrt{2ah} \\\\ v = \sqrt{2 \times 3.27 \times 0.6} \\\\ v = 1.98 \ m/s[/tex]

The speed of the mass m2 just before it hits the floor when there is friction is calculated as follows;

[tex]v = \sqrt{2ah} \\\\ v = \sqrt{2 \times 1.96 \times 0.6} \\\\ v = 1.53 \ m/s[/tex]

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Answer:

Mary

Explanation:

because if you look at her line, her distance is exceedin the others faster and higher. She has swam 2,200 m in 30 minutes while the other girls only swam 1,800 m and 1,200 m.

hence, Mary swam the most distance after 30 min.

have a nice night! :)

Answer: If the science lines up with the bible, then yes it is true.

If it doesn't, then it did not happen

Step by step answer:

Use this formula: [tex]v^{2}-v_{0}^{2} = 2aS[/tex]

Maximum speed: [tex]v = 33 m/s[/tex]

Started at rest: [tex]v_{0}=0[/tex]

Constant acceleration: [tex]a = 3.0 m/s^{2}[/tex]

So we can calculate [tex]S = \frac{33^{2}-0^{2} }{(2)(3.0)} =181.5m[/tex].

Answer:

the suppose the widtge frist is 6.5 inches and the lenth of your arm is 320 and 500

Based on the data provided, the initial, final and change in momentum of the carts are as follows:

Cart 1: initial = 1 kgm/s; final = 2.8 kgm/s; change = 1.8 kgm/s

Cart 2: initial = 6 kgm/s; final = 4.2 kgm/s; change = 1.8 kgm/s

Cart system: initial = 0 kgm/s; final = 7 kgm/s; change = 7 kgm/s

Momentum of a body is the product of its mass and velocity.

The law of conservation of momentum states that in a system of colliding bodies, the momentum is conserved provided there is no external force acting on the system.

mass = 0.5 kg, v = 2 m/s

momentum before collision = 0.5 * 2 = 1 kgm/s

mass = 0.75 kg, velocity = 8 m/s

momentum before collision = 0.75 kg * 8 m/s = 6 kgm/s

After collision, the stick together and move with a common velocity

From the principle of conservation of momentum;

m₁u₁ + m₂u₂ = (m₁ + m₂) * v

where v is common velocity

1 kgm/s + 6 kgm/s = (0.5 + 0.75) kg * v

v = 7/1.25 m/s

v = 5.6 m/s

Momentum = 0.5 kg * 5.6 m/s

Momentum = 2.8 kgm/s

Momentum = 0.75 kg * 5.6 m/s

Momentum = 4.2 kgm/s

2.8 kgm/s - 1 kgm/s

change in momentum = 1.8 kgm/s

4.2 kgm/s - 6 kgm/s

change in momentum = -1.8 kgm/s

Momentum of cart system before collision = 0

Momentum after collision = (0.5 + 0.75) kg * 5.6 m/s = 7 kgm/s

Change in momentum = 7 kgm/s - 0 = 7 kgm/s

Therefore, the initial, final and change in momentum of the carts are as follows:

Cart 1: initial = 1 kgm/s; final = 2.8 kgm/s; change = 1.8 kgm/s

Cart 2: initial = 6 kgm/s; final = 4.2 kgm/s; change = 1.8 kgm/s

Cart system: initial = 0 kgm/s; final = 7 kgm/s; change = 7 kgm/s

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