It's your first day on the job without your mentor as a member of mission control.
Your shift begins and your counterpart, who happens to be in their 10th year in this
position, speaks to you for about 45 minutes as they end their shift and you begin
yours. What were they talking to you about?

getting you up to date with recent mission developments

sanitation protocol that all employees must memorize

connecting you with each of the astronauts in space at the time

how to change settings for each system that's in use

(a) The electric field strength at a point 1.00 cm to the left of the middle is  2.0 x 10⁷ N/C.

(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)

E = +2.0 x 10⁷ N/C

F = Eq

F = 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = -94.4 N

Thus, the direction of the force will be towards the left.

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Answer:

Anticlockwise directions

Please mark me Brainliest to help me

Answer:

100rad because it angular velocity

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Mass of the astronaut, [tex]m_a[/tex] = 126 kg

Speed he acquires, [tex]v_{a}[/tex]  = 2.70 m/s

Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

[tex]P_f = m_av_a + m_cv_c[/tex]

[tex]P_I[/tex] = 0

[tex]m_av_a + m_cv_c = 0[/tex]

[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]

   [tex]= \frac{126* 2.70}{1800}[/tex]

   [tex]= - 0.189[/tex] m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     [tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     = [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

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The height of the light house is 17.86m.

To find the answer, we have to know about the simple pendulum.

                       [tex]T=2\pi \sqrt{\frac{l}{g} } \\[/tex]

where, g is acceleration due to gravity, and l is the length of the pendulum.

              [tex]h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m[/tex]

Thus, we can conclude that, the height of the light house is 17.86m.

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A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then

(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s

(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.

c) The speed of the 0.950- kg puck after the collision is 2.82 m/s

d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.

Mass of ice puck, m₁ = 0.350 kg

Mass of another puck, m₂ = 0.950 kg

Velocity of ice puck, v₁ = 5.22 m/s

Velocity of another puck, v₂ = 0 m/s

[tex]v^{'}_1= ?[/tex]

[tex]v^{'}_2= ?[/tex]

[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]

[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]

[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]

Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.

[tex]v^{'}_2 = v^{'} + v_1[/tex]

    [tex]= -2.40+5.22[/tex]

    [tex]= 2.82 m/s[/tex]

Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.

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Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum

Faraday's law states that a changing magnetic field through an area or equivalently, a changing area with constant field will cause a voltage.

So therefore, Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum

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The orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

The law of gravitation states that the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between the masses. Mathematically;

F = GMm/r²

where

M and m are the mass of ice cube and

Recall that;

s = Gm1/r^2

Also;

F = sm²

Substitute to have;

s = m²/F

For the centripetal acceleration

a = v²/r

Such that;

v²/r = Gm/r²

v² = Gm/r

v = √Gm/r

Substitute the given parameters into the formula to have:

V =  √6.67×10^-11 *  5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s = 3.56 * 10^6 m/s

Therefore the orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

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The height of the water level in the water tank is 0.011 m.

t = (2× 1.72)/9.8 = 0.35 s

Velocity along x direction= x × t

= 1.33 × 0.35

= 0.46 m/s

As per Bernoulli's theorem, 1/2 × ρ × v² = ρ × g× h

=> h = v²/2g = 0.46²/(2×9.8)

= 0.011 m

Thus, we can conclude that the height of water level is 0.011m.

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The required orbital speed of the ice cube is 355,358m/s

The force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. This can be expressed mathematically as;

Fr = GMm/r²

The distance is calculated as;

s = Gm/r²

Solving both equation, we will have:

v²/r = Gm/r²

v² = Gm/r

Take the square root of both sides

v = √Gm/r

Solve the required orbital speed

V =  √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s

Hence the required orbital speed of the ice cube is 355,358m/s

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Answer: 56.44°

Explanation:

Given:

Converting to SI Units (m/s):

= (1.2 mach)(340 ms^-1 / 1 Mach)

u  = 408 m/s

Find:

We have, sinθ = speed of sound / speed of object

               sinθ = v / u

                   θ = sin^-1 (v / u)  

                      = sin^-1 (340 / 408)

                   θ = 56.44°

When a piece of elastic material between two pieces of wood is being stretched beyond its limit, the elastic material or the object does not return to its original length when the force is removed

However, in physics, the type of stress applied when an elastic material is stretched is tensile stress.

Recall:

Stess is defined as force per unit area

Mathematically; Stress = F/A

Elasticity can be defined as the ability of a deformed elastic material or body to return to its original size and shape when the forces causing the deformation are removed.

So therefore, when a piece of elastic material between two pieces of wood is being stretched beyond its limit, the elastic material or the object does not return to its original length when the force is removed

Complete question:

What happens when a piece of elastic material between two pieces of wood is being stretched beyond its limit?

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The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

a)There is this impulse-momentum change equation.

[tex]where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.[/tex]

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

[tex]$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$[/tex]

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

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The option that is the correct one concerning the uncontrolled burn phase is:

Uncontrolled Combustion is known to be the the time and place in which a kind of an ignition will stop and it is said to be never  fixed by anything in regards to the compression ignition engine as seen in SI engines.

Note that the four Stages of combustion  are:

1.     Pre-flame combustion

2.     Uncontrolled combustion

3.     Controlled combustion and

4.     After burning

Hence, The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned as all the fuel need to burn out.

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Answer:

Gravity exist between two objects that have mass

The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

The resistance of a wire is the opposition to the flow of current. It depends on the following;

Given that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A

The given parameters are;

The formula to use to get R will be

R = ρL / A

Substitute all the necessary parameters into the formula

R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]

R = 8.6 x [tex]10^{-3}[/tex]  Ω

From Ohm's law, V = IR

Substitute all the necessary parameters into the formula

V = 0.2 x 8.6 × [tex]10^{-3}[/tex]

V = 1.72 x [tex]10^{-3}[/tex] V

Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

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Answer:

s^ -1    ( or    1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec   or   m/sec    

so b would have units of   s^-1

(or perhaps a more general term would be   time^-1)

Assuming point K is held in equilibrium, by Newton's second law we have

• net horizontal force

[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]

• net vertical force

[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]

where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies

[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]

Solve the first equation for [tex]F_C[/tex].

[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)

Substitute this into the second equation and solve for [tex]F_B[/tex].

[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)

[tex]F_B = \dfrac57 F_A[/tex]

[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]

The correct answer is 5828.675 J.

Given combined mass 4kg and mass of bullet 150gm=0.150kg.

Total mass= 4+0.150=4.150kg

Velocity=53 m/s

Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J

Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.

A body's translational kinetic energy is equal to  [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.

a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision ​

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Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

Mass of piece B = Mb

Ma×Va = Mb×Vb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

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Answer:C is the answer

Explanation:

It is the most reasonable and the answer that makes most sense

Answer:

The angle at which Earth's axis tilts changes.

Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.

An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.

An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.

Mathematically, the average velocity for this data set would be calculated by using this formula:

Average = [F(v)]/n

Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5

Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:

Substituting the parameters into the formula, we have:

Vavg = [300 + 450 + 500 + 250 + 300)/5

Vavg = 1800/5

Vavg = 360 m/s.

Next, we would calculate the percent difference by using this formula:

[tex]Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100[/tex]

Percent difference = [360 - 343]/360 × 100

Percent difference = 17/360 × 100

Percent difference = 0.0472 × 100

Percent difference = 4.72%.

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Answer:  It simply means that a freely falling object would increase its velocity by 9.8 m/s per second

Answer:

Hello!

"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that

That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.

The energy needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3 will be 1.93 *10^-18J and 12.09 eV.

The following can be deduced:

Energy of electron in hydrogen atom is

En = -13.6 /n2 eV

where n is principal quantum number of orbit.

Energy of electron in first orbit = E1 = -13.6 / 12 = - 13.6eV

Energy of electron in third orbit = E3 = -13.6 /32 = -1.51 eV

Energy required to move an electron fromfirst to thirdorbit ΔE = E3- E1

ΔE = -1.51 - ( 13.6) = 12.09 eV

Energy in Joule = 12.09 *l/× 1.6 × 10^-19 = 1.93 × 10^-18 J.

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Complete question:

How much energy is needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3? Give theanswer (a) in joules and (b) in eV.

The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.

We need to be aware of the Bernoulli principle in order to determine the solution.

                        [tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex]    We substitute d for to represent density.

                         [tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]

                      [tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]    

                       [tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]

Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.

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(a) The angular acceleration will be 26.035 rev/[tex]min^{2}[/tex].

(b) The final angular velocity is expected to be 33.846 rev/min.

Given.

t=1.3 min, Θ=22 rev, [tex]ω_{i}[/tex]=0

We know, Θ= [tex]ω_{i}[/tex]t+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]t^{2}[/tex]

22=0+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]1.3^{2}[/tex]

[tex]\alpha[/tex]=26.035 rev/[tex]min^{2}[/tex]

[tex]ω_{f} =ω_{i}+\alpha t[/tex]=0+26.035*1.3=33.846 rev/min

An object's rate of change in angular position or orientation over time is depicted by its angular velocity, rotational velocity, or both ( or ), also known as the angular frequency vector (i.e. how quickly an object rotates or revolves relative to a point or axis). The direction of the pseudovector is normal to the instantaneous plane of rotation or angular displacement, and its magnitude denotes the angular speed, or the rate at which the item rotates or revolves. It is customary to use the right-hand rule to specify the direction of angular motion. A general definition of angular velocity is "angle per unit time" (angle replacing distance from linear velocity with time in common). Radians per second is how angles are measured in the SI.

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The frictional force between the crate and the ramp is determined as 35.2 N.

Apply the principle of conservation energy to calculate the change in the energy of the crate.

Change in energy of the crate = energy loss to friction

P.Ei - K.Ef = E

mgh - ¹/₂mv² = E

where;

(30)(9.8)(2) - (0.5)(30)(2²) = E

528 J = E

E = Fd

where;

F = E/d

F = (528)/(15)

F = 35.2 N

Thus, the frictional force between the crate and the ramp is determined as 35.2 N.

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The maximum force constant of the spring Kmax is 2337.9 N/m.

The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.

Force constant is denoted by K and its unit is N/m.

Where;

K = spring constant

x = displacement

The work done by the spring is given below as follows:

Work done = Fx/2

Kinetic Energy = mv²/2

Force on an inclined plane = mgsinθ

Total force, F = mgsinθ + frictional force

F = 1390 * sin 22° + 515

F = 1035.7 N

Work done = change in KE

Fx/2 = mv²/2

Fx = mv²

m  = 1390/9.81 = 141.692

Solving for x;

x = mv²/F

x = 141.692 * 1.8²/1035.7

x = 0.443 m

The maximum force constant of the spring Kmax = 1035.7/0.443

Kmax = 2337.9 N/m

In conclusion, the maximum force constant of the spring  is the ratio of the total force and displacement.

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Note that the complete question is given below:

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria

Answer:

453 gm

Explanation:

Immersed objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm  in air