One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)

[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]f_{s}[/tex] - Static friction force, in newtons.

[tex]f_{k}[/tex] - Kinetic friction force, in newtons.

[tex]m[/tex] - Mass, in kilograms.

[tex]g[/tex] - Gravitational constant, in meters per square second.

If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:

[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{s} = 0.273[/tex]

[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.181[/tex]

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Answer:

The answer is below

Explanation:

Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.

a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.

Hence lead will have the highest temperature since it has the least specific heat capacity.

b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.

Hence aluminum will have the most heat since it has the highest specific heat capacity.

Answer:

Action and reaction are equal in magnitude and opposite in direction but they do not balance each other because they act on different objects so they don't cancel each other out.

hope this will help you more

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

1 metre per second = 2.237 miles per hour

90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour

90.1 m/s = 201.5537 mph

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

Answer:

Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad

Explanation:

GIven;

Wavelength of manganese vapor laser beam = 534 nm = 534 x 10⁻⁹ m

Diameter =  1.19 mm = 1.19 x 10⁻³ m

Find:

Minimum angular spread (in rad)

Computation:

Minimum angular spread (in rad) = 1.22[Wavelength / Diameter]

Minimum angular spread (in rad) = 1.222[(534 x 10⁻⁹) / (1.19 x 10⁻³)]

Minimum angular spread (in rad) = 2[448.73 x 10⁻⁶]

Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad

Answer:

The force on the second piston is 5246.5 N .

Explanation:

Area of first piston, a = 3.002 cm^2

Area of second piston, A = 315 cm^2

Force on first piston, f = 50 N

let the force of the second piston is F.

According to the Pascal's law

[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]

Answer:

a)     a = 27.44 m / s²,  b) a = 5.39 m / s², c)  a = 156.8 m / s², cabinet maximum acceleration does not change

Explanation:

a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.

Let's apply Newton's second law

we set a regency hiss where the x axis is in the direction of movement of the truck

Y axis y

        N- W = 0

        N = W = m g

X axis

       2fr = m a

the expression for the friction force is

      fr = μ N

      fr = μ m g

we substitute

      2 μ m g = m /2   a

     a = 4 μ g

      a = 4 0.7 9.8

      a = 27.44 m / s²

b) let's look for the maximum acceleration that can be applied to the cabinet

       fr = m a

       μ N = ma

       μ m g = m a

       a = μ g

       a = 0.55  9.8

       a = 5.39 m / s²

as the acceleration of the platform is greater than this acceleration the cabinet must slip

c) the friction force is in the four wheels as well

With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all

applying Newton's second law

         4 fr = (m/4) a

         16 mg = (m) a

         a = 16 g

         a = 16 9.8

         a = 156.8 m / s²

cabinet maximum acceleration does not change

[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]

Explanation:

Given:

[tex]\lambda = 4.96×10^3 s[/tex]

[tex]A_0 = 3.21x10^{17}[/tex] nuclei

t = 1.98×10^4 s

[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]

[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]

[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]

Answer:

105.62°F

Explanation:

When the body temperature having fever is measured to be 40.9 on a scale then it must be a Celsius scale thermometer because 37°C is the normal temperature of a healthy human. In case of fever the given temperature is measured on a standard Celsius scale.

The relation between Fahrenheit and Celsius scale is:

[tex]\frac{C}{5}=\frac{F-32}{9}[/tex]

[tex]F=\frac{9C}{5} +32[/tex]

[tex]F=105.62^{o}F[/tex]

It is a high fever and an immediate medical help must be taken.

Complete question:

A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answer:

The tension on the rope is 1.95 N

Explanation:

The general equation of a progressive wave is given as;

[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]

Compare the given equation with the general equation of wave, the following parameters will be deduced.

A = 0.75 cm

k = 0.400π cm⁻¹

ω = 250π s⁻¹

The frequency of the wave is calculated as;

ω = 2πf

2πf = 250π

2f = 250

f = 250/2

f = 125 Hz

The wavelength of the wave is calculated as;

[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]

The velocity of the wave is calculated as;

v = fλ

v = 125 x 0.05

v = 6.25 m/s

The tension on the rope is calculated as;

[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]

Therefore, the tension on the rope is 1.95 N

Answer:

a.v=u+v/2

a.v=s/t

combining two equation we get,

u+v/2=s/t

(u+v)t/2=s

(u+v)t/2=s

{u+(u+at)}t/2=s

(u+u+at)t/2=s

(2u+at)t/2=s

2ut+at^2/2=s

2ut/2+at^2/2=s

UT +1/2at^2=s

proved

a=v-u/t

at=v-u

u+at=v

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

Answer:

(b)

Explanation:

The voltage always lags the current by 90°, regardless of the frequency.

(a) As it gets compressed by a distance x, the spring does

W = - 1/2 (52.1 N/m) x ²

of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so

W = ∆K

- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.118 m

(b) Taking friction into account, the only difference is that more work is done on the object.

By Newton's second law, the net vertical force on the object is

∑ F = n - mg = 0

where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives

n = mg = 2.45 N

and from this we get the magnitude of kinetic friction,

f = µn = 0.120 (2.45 N) = 0.294 N

Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:

W (friction) = - (0.294 N) x

W (spring) = - 1/2 (52.1 N/m) x ²

==>   W (total) = W (friction) + W (spring)

Solve for x :

- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.112 m

For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:

a) The spring's maximum compression when the track is frictionless is 0.118 m.

b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.

a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex]  (1)

Where:

[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg

[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s

k: is the spring constant = 52.1 N/m

x: is the distance of compression

After solving equation (1) for x, we have:

[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]

Hence, the spring's maximum compression is 0.118 m.

b) When the track is not frictionless, we can calculate the spring's compression by work definition:

[tex] W = \Delta E = E_{f} - E_{i} [/tex]

[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex]   (2)

Work is also equal to:

[tex] W = F*d = F*x [/tex]     (3)

Where:  

F: is the force

d: is the displacement = x (distance of spring's compression)  

The force acting on the object is given by the friction force:

[tex] F = -\mu N = -\mu m_{o}g [/tex]   (4)

Where:

N: is the normal force = m₀g

μ: is the coefficient of kinetic friction = 0.120

g: is the acceleration due to gravity = 9.81 m/s²

The minus sign is because the friction force is in the opposite direction of motion.

After entering equations (3) and (4) into (2), we have:

[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]

[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]

[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]        

Solving the above quadratic equation for x

[tex] x = 0.112 m [/tex]  

Therefore, the spring's compression is 0.112 m when the track is not frictionless.

Read more here:

I hope it helps you!  

Answer:

Liquid volume is usually measured using either a graduated cylinder or a buret. As the name implies, a graduated cylinder is a cylindrical glass or plastic tube sealed at one end, with a calibrated scale etched (or marked) on the outside wall.

Answer:

2) The bending of a light wave as it passes at an angle from one medium to another .

Hope it is helpful to you ☺️

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².

Answer:

D

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex]  Filling in:

[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

Complete question is;

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly

downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

Answer:

Power ≈ 600,000 W

Explanation:

We are given;

Frequency; f = 2400 Hz

height of the oven cavity; h = 25 cm = 0.25 m

base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²

total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J

We want to find the power output and we know that formula for power is;

P = workdone/time taken

Formula for time here is;

t = h/c

Where c is speed of light = 3 × 10^(8) m/s

Thus;

t = 0.25/(3 × 10^(8))

t = 8.333 × 10^(-10) s

Thus;

Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))

Power ≈ 600,000 W

Answer:

The angular acceleration is 4.5 rad/s^2.

Explanation:

Acceleration, a = 26.5 m/s2

length, L = 5.89 m

The angular acceleration is

[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Answer:

because minerals can be gotten from the bottom

Explanation:

it's self explanatory

Answer:

150 N

Explanation:

Given that,

We are asked to calculate force required.

[tex]\longrightarrow[/tex] F = ma

[tex]\longrightarrow[/tex] F = (50 × 3) N

[tex]\longrightarrow[/tex] F = 150 N

Answer:

wire 66.0 cm long carries a 0.750 A current in the positive direction of an x axis through a magnetic field $$\vec { B } = ( 3.00 m T ) \hat { j } ...

Top answer · 1 vote

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

Answer:

Average Velocity = 3.63 m/s

Explanation:

First, we will calculate the total displacement of the quarterback, taking forward direction as positive:

Total Displacement = 15 m - 3 m + 24 m = 36 m

Now, we will calculate the total time taken for this displacement:

Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s

Therefore, the average velocity will be:

[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]

Average Velocity = 3.63 m/s