what is the static friction force and kinetic friction for the image

Answer:

Answer:

Speed of the wave in the string will be 3.2 m/sec

Explanation:

We have given frequency in the string fixed at both ends is 80 Hz

Distance between adjacent antipodes is 20 cm

We know that distance between two adjacent anti nodes is equal to half of the wavelength

So \frac{\lambda }{2}=20cm

2

λ

=20cm

\lambda =40cmλ=40cm

We have to find the speed of the wave in the string

Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec

So speed of the wave in the string will be 3.2 m/sec

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

Since we're finding the velocity we make v the subject

That's

[tex]v = \sqrt{ \frac{2k}{m} } \\ [/tex]

From the question

k = 5 J

m = 0.1 kg

We have

[tex]v = \sqrt{ \frac{2 \times 5}{0.1} } = \sqrt{ \frac{10}{0.1} } = \sqrt{100} \\ = 10[/tex]

We have the final answer as

Hope this helps you

This question involves the concepts of orbital velocity and orbital radius.

The orbital velocity of ISS must be "7660.25 m/s".

The orbital velocity of the ISS can be given by the following formula:

[tex]v=\sqrt{\frac{GM}{R}}[/tex]

where,

v = orbital velocity = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m

R = 67.86 x 10⁵ m

Therefore,

[tex]v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}[/tex]

v = 7660.25 m/s

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Answer:

electricity because it has more percentage nd energy

Explanation:

mark me brainliest pl

Answer: pinkish- white

Purple

Explanation:

Answer:

B,A,A

Explanation:

Answer:

Other guy is correct b,a,a

Explanation:

Answer:

F(10)=mass(327)x acceleration(m/s)

Explanation:

From Newton's second law of motion, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]

Given that a 25 kg crate is pushed horizontally with a force of 70N and the coefficient of friction is 0.20

To calculate the acceleration of the crate, We will consider the net force acting on the crate. The net force will be sum of the applied force F and the frictional force [tex]F_{r}[/tex].

Let us first calculate the frictional force [tex]F_{r}[/tex]. That is,

[tex]F_{r}[/tex] =- μN

[tex]F_{r}[/tex] = μmg

[tex]F_{r}[/tex] = 0.2 x 25 x 9.8

[tex]F_{r}[/tex] = 49N

Using Newton's second law,

F = ma

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

70 - 49 =  25 a

21 = 25a

a = 21 / 25

a = 0.84 m/[tex]s^{2}[/tex]

Therefore, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]

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Answer:

They must be supported by evidence.

Explanation:

Every scientific theory or scientifc claim must have scientific evidence.

[tex]\\ \sf\Rrightarrow s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \sf\Rrightarrow s=0(19)+\dfrac{1}{2}a(19)^2[/tex]

[tex]\\ \sf\Rrightarrow 475=\dfrac{361a}{2}[/tex]

[tex]\\ \sf\Rrightarrow 950=361a[/tex]

[tex]\\ \sf\Rrightarrow a\approx 2.3m/s^2[/tex]

Answer: I don't understand

Explanation:

study and pay attention

Answer:

0/10

Explanation:

_--------------------

Answer:

No. -infinity/10

Explanation:

Ew. Just why, like no.

Answer:

Explanation:

See the attachment for the details.  A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times.  The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.  

The individual driving times and distances are summed to provide:

(a) How far did he travel?

103 km

(b) What was his average speed?

12.88 km/h

(c) What was his displacement?

73.82 km

(d) What was his average velocity?

9.228 km/h

The total displacement of the toy car at the given positions is 0.

The given parameters;

The total displacement of the car is calculated as follows;

[tex]\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0[/tex]

Thus, the total displacement of the toy car at the given positions is 0.

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Answer:0

Explanation:

Answer:

“The only true wisdom is in knowing you know nothing.” ― Socrates. Socrates had it right. Remain humble and appreciate the fact that you don't know everything. ... Thankfully, one's search for knowledge has no end as there is always something new to learn, giving one's life meaning and purpose.

Explanation:

Answer:

knowing wisdom cool

Explanation:

lol

Answer:

Hey Tyere!

Refer to the attachments....

Answer:

Answer to the question is 4 / 5

Answer:

Explanation:

Momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity.

The period of an oscillating body is the smallest interval of time in which a system undergoing oscillation returns to the state it was in at a time arbitrarily chosen as the beginning of the oscillation.

Friction is a force between two surfaces that are sliding, or trying to slide, across each other.  Friction always slows a moving object down. The amount of friction depends on the materials from which the two surfaces are made. The rougher the surface, the more friction is produced.

From Kepler's third law, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C

Given that an astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years.

According to Kepler's third law,

[tex]T^{2} \alpha r^{3}[/tex]

Where

T = Period ( in earth years) = time to complete one orbit

r = Length of the semi major axis in Astronomical unit.

[tex]T^{2}[/tex] = [tex]\frac{4\pi ^{2} }{GM} * r^{3}[/tex]

convert years to seconds

105 x 365 day x 24 hours x 3600 s

T = 3311280000 seconds

Mass of the sun M = 1.989 × 10^30 kg

G = 6.67 x [tex]10^{-11}[/tex]N m^2/kg^2

Substitute all the parameters into the formula

[tex]T^{2}[/tex] = 1.096 x [tex]10^{19}[/tex] = [tex]\frac{4\pi ^{2} }{6.67 * 10^{-11} * 1.989 * 10^{30} } * r^{3}[/tex]

1.096 x [tex]10^{19}[/tex] = 2.976 x [tex]10^{-19}[/tex] [tex]r^{3}[/tex]

[tex]r^{3}[/tex] = 1.096 x [tex]10^{19}[/tex] / 2.976 x [tex]10^{-19}[/tex]

[tex]r^{3}[/tex] = 3.68 x [tex]10^{37}[/tex]

r = [tex]\sqrt[3]{3.68 * 10^{37} }[/tex]

r = 3.33 x [tex]10^{12}[/tex] m

1 AU = 1.5 x [tex]10^{11}[/tex] m

r = 3.33 x [tex]10^{12}[/tex] / 1.5 x [tex]10^{11}[/tex]

r = 22.18 AU

Therefore, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C

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Answer:

C. 22.3 AU

Explanation:

Not only is the above an unnecessarily complicated answer, it's not even fully correct, and definitely not what they want you to do.

T^2 = s^3, where T = orbital period and s = semi-major axis length.

Substitute T and you get 105^2 = s^3. Solve for s.

11025 = s^3

3√11025 = s

22.25663649 = s

Therefore, the answer is C. 22.3 AU

An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.

The magnitude of an earthquake is a number that characterizes the relative size of an earthquake and is based on the measurement of the maximum motion recorded by a seismograph.

If the magnitude of the earthquake is M = 5.5, we can calculate the amount of energy released (E) using the following expression.

[tex]M = \frac{2}{3} log(\frac{E}{E_0} )\\\\E = E_0 \times 10^{\frac{3}{2}M } = 10^{4.4} \times 10^{\frac{3}{2}(5.5) } = 4.5 \times 10^{12} J[/tex]

where,

An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.

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Answer:

2 * 10^4

Explanation:

(3 * 10^4)(4 * 10^4) / (6 * 10^4)

= (12 * 10^8)/(6 * 10^4)

= 2 * 10^4

Answer:

The answer is 1200

Explanation:

Answer:

a planet

Explanation:

If planetesimals, objects that combined to form protoplanets, were to collide head on in the early solar nebula, they would shatter to the point where they would be unable to grow into a larger size, a protoplanet. A protoplanet is a massive object that will eventually become a planet.

[tex] = \sf \frac{1}{2} m {v}^{2} [/tex]

[tex] \sf = \frac{2.4J}{9.6J} \\ \sf= \frac{24}{96} = \frac{1}{4} \\ \sf = 1 : 4[/tex]

Answer:

The kinetic energies of the bullets are 2.4 J and 9.6 J.

The bullet having greater velocity has more kinetic energy.

The ratio of their kinetic energies is 1 : 4.

Hope you could get an idea from here

Doubt clarification - use comment section.

The kinetic energy of the two bullets are 2.4 J and 9.6 J respectively.

The ratio of the kinetic energy of the bullets is 1:4.

The kinetic energy of an object is the energy possessed by the object due to its motion.

The kinetic energy of the two bullets is calculated as follows;

[tex]K.E_1 = \frac{1}{2} \times 0.003 \times 40^2 = 2.4 \ J\\\\K.E_2 = \frac{1}{2} \times 0.003 \times 80^2 = 9.6 \ J[/tex]

Ratio of the kinetic energy of the bullets is calculated as follows;

[tex]K.E_1 : K.E_2 = 2.4: 9.6 \ = \ 1: 4[/tex]

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The acceleration of the bicycle is 0.12 m/[tex]s^{2}[/tex].

We know that acceleration is the change in velocity divided by the change in time. In this case, the bicycle is accelerating from rest to a velocity of 6 m/s. The change in time is the time it takes the bicycle to travel 50 m.

We can use the following equation to find the acceleration:

a = (v2 - v1) / t

where:

a is the acceleration

v2 is the final velocity (6 m/s)

v1 is the initial velocity (0 m/s)

t is the time

Plugging in the values from the problem, we get:

a = (6 m/s - 0 m/s) / 50 s

= 0.12 m/[tex]s^{2}[/tex].

Therefore, the acceleration of the bicycle is 0.12 m/[tex]s^{2}[/tex].

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Answer:

It can be measure by grams

Answer:

c. Only neutrons.

hope it helps :)

Answer:

The magnitude of the student's displacement is less than the distance she walked.

Walking:  11 Blocks

Displacement:  8.54 Blocks

Explanation:

See the attached diagram.  The unit of length is blocks.  We can add the actual blocks walked as shown.  She walked a total of 11 blocks.

Her displacement is the distance measured directly from where she started (line A).  Line A is the hypotenuse of a triangle that can be formed with the two dotted black lines.  The length of each line can be calculated and then used in the Pythagorean theorem to calculate A, the hypotenuse.

That result is 8.54 Blocks, a shorter distance, once she earns her wings.

If a student on her way to school walks four blockers east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than the total distance walked by her.

An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard.

As given in the problem if a  student on her way to school walks four blockers east, three blocks north, and another four blocks east,

The total distance walked by the student = 11 blocks

Displacement of the student from home = √(8² + 3²)

                                                                   = 8.5 blocks

The total displacement by the student would be less than the

Thus, the magnitude of her displacement from home is less than the distance.

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Answer:

all of the particles are at rest (v = 0)

Explanation:

Therefore, since nothing is moving, the total momentum of the system must also be zero.

As a system consisting of two particles is known to have zero total momentum, the kinetic energy of the system will be zero as well.

Momentum is the product of a particle's mass and velocity. Momentum is a vector quantity, which means it has a magnitude as well as a direction.

According to Isaac Newton's second law of motion, the time rate of change of momentum equals the force acting on the particle.

Momentum describes the relationship between an object's mass, velocity, and direction.

Any change in momentum generates force. As a result, a change in momentum is used to calculate the force acting on the object.

Because the total kinetic energy is zero, all of the particles are at rest (v = 0). As a result, because nothing is moving, the system's total momentum must also be zero.

Thus, the statement is true.

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