A muon is produced by a collision between a cosmic ray and an oxygen nucleus in the upper atmosphere at an altitude of 50 . It travels vertically downward to the surface of the Earth where it arrives with a total energy of 178 . The rest energy of a muon is 105.7 . What is the kinetic energy of a muon at the surface

Answer:

The question is incomplete,below is the complete question

"Calculate the number of atoms contained in a cylinder (1μm radius and 1μm deep)of (a) magnesium (b) lead."

Answer:

a. 1.35*10^{11} atoms

b. 1.03*10^{11} atoms

Explanation:

First, we determine the volume of the magnesium in the cylinder container

using the volume of a cylinder

[tex]V=\pi r^{2}h\\ r=10^{-6}m\\ h=10^{-6}m\\V=\pi *10^{-6*2}*10^{-6}\\V=\pi *10^{-18}\\V=3.14*10^{-18}m^{3}\\[/tex]

a. Next we determine the mass of the magnesium ,

using the density=mass/volume

since density of a magnesium

[tex]the density of magnesium =1.738*10^{3}kg/m^{3} \\mass=density * volume \\mass=1.738*10^{3}*3.14*10^{-18}\\mass=5.46*10^{-15}kg\\ \\mass=5.46*10^{-12}g\\[/tex]

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

[tex]mole=5.46*10^{-12}/ 24.305\\mole=0.225*10^{-12}mol\\[/tex]

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

[tex]number of atoms = 0.225*10^{-12}*6.02*10^{23}\\number of atoms =1.35*10^{11} atoms[/tex]

b. for he lead, we determine the mass of the lead  ,

using the density=mass/volume

since density of a magnesium

[tex]the density of lead =11.34*10^{3}kg/m^{3} \\mass=density * volume \\mass=11.34*10^{3}*3.14*10^{-18}\\mass=35.60*10^{-15}kg\\ \\mass=35.60*10^{-12}g\\[/tex]

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

[tex]mole=35.60*10^{-12}/ 207.2\\mole=0.1718*10^{-12}mol\\[/tex]

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

[tex]number of atoms = 0.1718*10^{-12}*6.02*10^{23}\\number of atoms =1.03*10^{11} atoms[/tex]

Answer:

19m/s

22.3 degrees

Explanation:

it is a case aof relative velocity.

the basic for relative velocity vector equation is :

V_b = V_j + V_(b/j)---------------1

V_b: ball velocity relative to ground

V_j : Jaun velocity relative to ground

V_(b/j): ball velocity relative to jaun

reference frame:

We take east and north as +ve x and +ve y

V_(b/j) = V_b - V_j

so for x-axis;

net x-component of V_(b/j) = 12 sin (37) + 0 = 7.22m/s

net y-component of V_(b/j) = 12 cos (37) + 8 = 17.6m/s

magnitude = ((7.22^2)+(17.6^2))^(0.5) = 19 m/s

*direction with respect to Jaun = angle between the vertical (North) and vector V_(b/j)

angle = arctan(7.22/17.6) = 22.3 degrees

Answer:

[tex]a=24.025\ m/s^2[/tex]    

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

[tex]a=15.5^2\times 0.1\ m/s^2[/tex]

[tex]a=24.025\ m/s^2[/tex]

Therefore the acceleration of the clay will be [tex]a=24.025\ m/s^2[/tex].

The question involves analyzing the projectile motion of a rock thrown from a building using physics principles like motion decomposition and energy conservation. It requires breaking down the initial velocity into horizontal and vertical components and applying kinematic equations to determine parameters such as max height, range, and flight time.

This question involves the principles of projectile motion and energy conservation in physics. When the man throws the rock at an angle of 28 degrees above the horizontal with an initial velocity of 26.0 m/s from a height of 14.0 meters, we need to analyze the horizontal and vertical components of the motion separately to determine various aspects of the rock's trajectory, such as its range, maximum height, and time of flight. However, since the specific request is missing in this context, we'll focus on the general approach to solving such problems.

To solve problems involving objects thrown at an angle, we first decompose the initial velocity into its horizontal (vx = v*cos(θ)) and vertical (vy = v*sin(θ)) components, where v is the magnitude of the initial velocity and θ is the angle of projection. The horizontal motion is uniform, meaning the velocity remains constant, whereas the vertical motion is affected by gravity, leading to acceleration in the opposite direction of the initial vertical velocity component.

Energy conservation or kinematic equations can be used to find specific details like maximum height reached, time of flight, and range. For example, the formula s = ut + 0.5at² can be applied where s is displacement, u is initial velocity, a is acceleration (gravity in the case of vertical motion), and t is time. Remember, acceleration due to gravity (a) is -9.8 m/s², indicating it acts downwards. Ignoring air resistance simplifies calculations by omitting drag force considerations.

The maximum horizontal distance is approximately 61.7 meters. This is determined using the projectile motion equations for horizontal distance with initial velocity, angle, and height given.

To find the maximum horizontal distance the rock travels, we can analyze the projectile motion. The initial velocity of 26.0 m/s is broken down into horizontal and vertical components. The horizontal component is [tex]\( v_x = v \cdot \cos(\theta) \)[/tex], where \( v \) is the magnitude of the velocity (26.0 m/s) and \( \theta \) is the angle (28.0 degrees). The vertical component is [tex]\( v_y = v \cdot \sin(\theta) \).[/tex]

The vertical motion is affected by gravity, and the time it takes for the rock to hit the ground can be calculated using the equation [tex]\( h = v_y \cdot t - \frac{1}{2} g t^2 \),[/tex] where \( h \) is the initial height (14.0 m), \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)), and \( t \) is the time of flight.

Using the quadratic formula to solve for \( t \), we find two solutions: one when the rock is at the initial height and one when it hits the ground. We use the positive solution for the time of flight to calculate the horizontal distance traveled using the equation [tex]\( d = v_x \cdot t \).[/tex]

Substituting the values, we find the maximum horizontal distance to be approximately 61.7 meters.

The question probably maybe: What is the maximum horizontal distance the rock travels before hitting the ground, given that a man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28.0 degrees above the horizontal, ignoring air resistance?

The question is incomplete but an analysis of the situation using  various useful physics concepts can still be made

Answer:

Answer:

Explanation:

Given

velocity of diver [tex]u=-10\ m/s[/tex] i.e. downward motion

acceleration due to gravity [tex]a=g=-9.8\ m/s^2[/tex]

time [tex]t=1.16\ s[/tex]

using equation of motion

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

[tex]y=(-10)\cdot 1.16-\frac{1}{2}(-9.8)(1.16)^2[/tex]

[tex]y=-11.6-6.593[/tex]

[tex]y=-18.19\ m[/tex]

I.e. in downward direction                    

The displacement of the diver during the last 1.16 seconds of her dive is -11.6 meters. The negative sign indicates a downward movement.

In physics, displacement is the overall change in position of an object. It is calculated by multiplying velocity and time. In this case, the velocity of the diver is -10.0 m/s (a negative sign indicating downward motion) and the time is 1.16 s. To find the displacement, multiply the velocity and the time: (-10.0 m/s) × (1.16 s) = -11.6 m. The minus sign still indicates downward direction, and it means that the diver moved 11.6 meters downward in the last 1.16 seconds of her dive.

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Answer:

E=[tex]\frac{KQ}{2\sqrt 2a^2}[/tex]

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

[tex]E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}[/tex]

Substitute x=a and R=a

Then, we get

[tex]E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}[/tex]

[tex]E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}[/tex]

[tex]E=\frac{KQa}{2\sqrt 2a^3}[/tex]

[tex]E=\frac{KQ}{2\sqrt 2a^2}[/tex]

Where K=[tex]9\times 10^9 Nm^2/C^2[/tex]

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=[tex]\frac{KQ}{2\sqrt 2a^2}[/tex]

The magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center is E = Q/[8√2πε₀a²]

The electric field due to a charged ring E is given by

E = Qz/4πε₀[√(z² + R²)]³ where

Given that for this ring R = a and z = a, substituting these values into E, the magnitude of the electric field at a is given by

E = Qz/4πε₀[√(z² + R²)]³

E = Qa/4πε₀[√(a² + a²)]³

E = Qa/4πε₀[√(2a²)]³

E = Qa/4πε₀[2√2a³]

E = Q/[8πε₀√2a²]

E = Q/[8√2πε₀a²]

So, the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center is E = Q/[8√2πε₀a²]

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Answer:

Explanation:

Given

Voltage of battery [tex]V=45\ V[/tex]

electric field(E) is given by  

[tex]E=|-\frac{dV}{dx}|[/tex]

i.e. Change in Potential over a distance x

(a)When Plate are separated by 3 cm apart

[tex]E=|\frac{45}{3\times 10^{-2}}|[/tex]

[tex]E=1500\ V/m[/tex]

(b)When Plates are 5 cm apart

[tex]E=|-\frac{45}{5\times 10^{-2}}|[/tex]

[tex]E=900\ V/m[/tex]

Force on electron is given by

[tex]F=charge\times Electric\ Field[/tex]

[tex]F=q\times E[/tex]

For case a

[tex]F=1.6\times 10^{-19}\times 1500[/tex]

[tex]F=24\times 10^{-17}\ N[/tex]

(b)[tex]F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N[/tex]

Answer:

[tex]a=-3\ m/s^2[/tex]

Explanation:

Second Newton's Law

It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by

[tex]F_n=m.a[/tex]

The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is

[tex]F_n=30\ N-60\ N=-30\ N[/tex]

Thus, the acceleration can be computed by

[tex]\displaystyle a=\frac{F_n}{m}=\frac{-30}{10}=-3\ m/s^2[/tex]

[tex]\boxed{\displaystyle a=-3\ m/s^2}[/tex]

The negative sign indicates the block is accelerated to the left

The block will experience an acceleration of 3.0 m/s² to the left, as calculated by using the net force (30 N leftward) divided by the mass of the block (10.0 kg).

To determine the acceleration of a 10.0 kg block on a smooth horizontal surface that is being acted upon by two forces, one must first find the net force acting on the block. In this scenario, there is a 30 N force acting to the right and a 60 N force acting to the left. The net force can be found by subtracting the smaller force from the larger one, which gives us a result of 60 N - 30 N = 30 N acting to the left.

Now, using Newton's second law of motion which states that F = ma (where F is the net force, m is the mass of the object, and a is the acceleration), we can solve for acceleration by rearranging the equation to a = F/m. Plugging the net force (F = 30 N) and the mass of the block (m = 10.0 kg) into the equation yields:

a = 30 N / 10.0 kg = 3.0 m/s²

The acceleration of the block will be 3.0 m/s², and since the net force is to the left, the acceleration will also be directed to the left.

Answer:

C)T

Explanation:

The period of a mass-spring system is:

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

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

Answer: The value of spring constant is 121.9 N/m

Explanation:

Force is defined as the mass multiplied by the acceleration of the object.

[tex]F=m\times g[/tex]

where,

F = force exerted on the object  = ?

m = mass of the object  = 0.46 kg

g = acceleration due to gravity = [tex]9.81m/s^2[/tex]

Putting values in above equation, we get:

[tex]F=0.46kg\times 9.81m/s^2=4.51N[/tex]

To calculate the spring constant, we use the equation:

[tex]F=k\times x[/tex]

where,

F = force exerted on the spring = 4.51 N

k = spring constant = ?

x = length of the spring = 3.7 cm = 0.037 m     (Conversion factor:  1 m = 100 cm)

Putting values in above equation, we get:

[tex]4.51N=k\times 0.037m\\\\k=\frac{4.51N}{0.037m}=121.9N/m[/tex]

Hence, the value of spring constant is 121.9 N/m

Answer:

[tex]P=1139.16384mmHg[/tex]

Explanation:

Given data

[tex]R=0.08206(\frac{L.Atm}{mol.K} )\\Density=2.697g/L\\Temperature=298K\\f.wt=44(g/mol)\\[/tex]

To find

Pressure

Solution

From Ideal gas law we know that

[tex]PV=nRT\\P=(nR\frac{T}{V} )=(R(\frac{mass}{f.wt} )(\frac{T}{V} ))\\P=R(\frac{mass}{volume}) (\frac{T}{f.wt} )=R(Density)(\frac{T}{f.wt} )[/tex]

Substitute the given values to find pressure

So

[tex]P=(0.08206\frac{L.Atm}{mol.K} )(2.697g/L)(298K)(44g/mol)^{-1}\\ P=1.4989Atm\\[/tex]

Convert Atm to mmHg

Multiply the pressure values by 760

So

[tex]P=1139.16384mmHg[/tex]

Answer:

1.15 rad/s²

Explanation:

given,

angular speed of turntable = 45 rpm

                       =[tex]45\times \dfrac{2\pi}{60}[/tex]

                       =[tex]4.71\ rad/s[/tex]

time, t = 4.10 s

initial angular speed = 0 rad/s

angular acceleration.

[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]

[tex]\alpha = \dfrac{4.71-0}{4.10}[/tex]

[tex]\alpha = 1.15\ rad/s^2[/tex]

Hence, the angular acceleration of the turntable is 1.15 rad/s²

The ball's acceleration is constant in magnitude and direction, from the instant it leaves your hand, until the instant it hits the ground, no matter what direction or speed you throw it.

It's the acceleration of gravity, on whatever planet you happen to be standing when you throw the ball.

Final answer:

The baseball's magnitude of acceleration is greater while being thrown than after it leaves the hand due to the additional force applied by the thrower, while in free fall, the ball is subject only to gravity. With air resistance, the ball takes longer to go up than to come back down.

Explanation:

The question pertains to the acceleration of a baseball when thrown straight up into the air. While being thrown, the ball experiences an acceleration greater than the acceleration due to gravity because of the force applied by the person's arm. After the ball leaves the hand, however, the only force acting on it is the force of gravity, which gives it a constant acceleration of approximately 9.81 m/s² downward, regardless of air resistance. In the absence of other forces, the magnitude of acceleration when the ball is in free fall is less than the acceleration imparted to the ball by the thrower's arm.

When air resistance is considered, it acts to slow down the ball as it rises and speeds up as it falls. Therefore, with air resistance, the time it takes for the ball to go up is greater than the time it takes to come back down, because air resistance removes kinetic energy from the ball on the way up, slowing it down more quickly than gravity alone would.

Answer:

[tex]g \approx 7.4 m/s^2[/tex]

Explanation:

Assuming the following data:

Heigth (m): 0 , 1, 2, 3, 4, 5

Time (s): 0.00, 0.54,0.73, 0.91, 1.01, 1.17

We know from kinematics that the height is given by the following expression:

[tex] h_f = h_i + v_i t + \frac{1}{2} g t^2[/tex]

Assuming for this case that the initial velocity is [tex]v_i[/tex] we can find a polynomial with degree 2 in order to have an estimation for the height with the time.

We can use excel for this and we can see the polynomial adjusted for the data given.

As we can see the best polynomial of degree 2 is given by:

[tex] h(x)= -0.0178 +0.066 x+ 3.6801 x^2[/tex]

For our case x = time and we can rewrite the expression like this

[tex] h(t)= -0.0178 +0.066 t+ 3.6801 t^2[/tex]

And if we are interested on the gravity we want on special the last term of this equation, we can set equal the following terms:

[tex] 3.6801 t^2 = \frac{1}{2} g t^2[/tex]

And solving for g we got:

[tex] 2*3.6801 = g= 7.36 \frac{m}{s^2}[/tex]

So then a good approximation for the gravity of the planet rounded to 2 significant figures is 7.4 m/s^2

To determine the free-fall acceleration on Planet X, you can use the formula: acceleration = 2 * height / fall time^2. Calculating the acceleration for each data point, and the uncertainty can be determined by finding the range of values.

a.

To determine the free-fall acceleration on Planet X, we can use the formula: acceleration = 2 * height / fall time^2. We can calculate the acceleration using the given data:

For height 1m, fall time 0.54s, the acceleration is 7.485 m/s^2

For height 2m, fall time 0.72s, the acceleration is 8.660 m/s^2

For height 3m, fall time 0.91s, the acceleration is 9.337 m/s^2

For height 4m, fall time 1.01s, the acceleration is 9.704 m/s^2

For height 5m, fall time 1.17s, the acceleration is 9.335 m/s^2

b.

To determine the uncertainty in the free-fall acceleration, we can calculate the range of values by subtracting the smallest acceleration from the largest acceleration. The smallest acceleration is 7.485 m/s^2 and the largest acceleration is 9.704 m/s^2. Therefore, the uncertainty in the free-fall acceleration is 2.219 m/s^2.

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Answer:

Explanation:

Given

acceleration is given by

[tex]a=-g-c_Dv^2[/tex]

where [tex]\ddot{y}=a[/tex]

[tex]\dot{y}=v[/tex]

Also acceleration is given by

[tex]a=v\frac{\mathrm{d} v}{\mathrm{d} s}[/tex]

[tex]ds=\frac{v}{a}dv[/tex]

[tex]\int ds=\int \frac{v}{-g-0.001v^2}dv[/tex]

[tex]\Rightarrow Let -g-0.001v^2=t[/tex]

[tex]-0.001\times 2vdv=dt[/tex]

[tex]vdv=-\frac{dt}{0.002}[/tex]

[tex]at\ v_0=50\ m/s,\ t=-g-0.001(50)^2[/tex]

[tex]t=-g-2.5[/tex]

at [tex]v=0,\ t=-g[/tex]

[tex]\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}[/tex]

[tex]\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}[/tex]

[tex]s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}[/tex]

[tex]s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})[/tex]

[tex]s=113.608\ m[/tex]

when air drag is neglected maximum height reached is

[tex]h=\frac{v_0^2}{2g}[/tex]

[tex]h=\frac{50^2}{2\times 9.8}[/tex]

[tex]h=127.55\ m[/tex]

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Answer:

a. by collisions and mergers of planetesimals.

Explanation:

Inner planets are planets within 1.5 AU distance from the sun. These are called terrestrial planets because they are somewhat similar to Earth, mainly made of rocks.

The main ingredient of these planets are solar nebula and interstellar dust condensation of which leads to formation of small rock particles. These particles come close to each other under in the influence of gravity and other forces. As the mass of the particles increase they form planetesimals, these planetesimals eventually merge to form planets.

The value of C if the object stops in 8.00 s is 0.625 m/s³.

The distance traveled by the object before stopping in 8 seconds is 40 m.

The given parameters;

The value of C is determined by using velocity equation as shown below;

[tex]\frac{dv}{dt} = -Ct\\\\dv = -Ctdt\\\\\int\limits^v_{v_0} \, dv = -\int\limits^t_{t_0} \, Ct \\\\v-v_0= -C[\frac{t^2}{2} ]^t_0\\\\v-v_0 = - \frac{1}{2} Ct^2\\\\0 - 20 = - \frac{1}{2}C(8)^2\\\\-20 = -32 C\\\\C = \frac{20}{32} = 0.625 \ m/s^3[/tex]

The acceleration of the object during 8 seconds is calculated as follows;

a = -Ct

a = -0.625(8)

a = -5 m/s²

The distance traveled by the object before stopping in 8 seconds is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + 2(-5)s\\\\0 = 400 - 10s\\\\10s = 400 \\\\s = \frac{400}{10} \\\\s = 40 \ m[/tex]

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A value of C that stops the object in 8 seconds is 1.25 m/s³. The object travels a total distance of 53.33 meters during this time.

An object is moving along the x-axis with an initial velocity of v₀x = 20.0 m/s and an acceleration of ax = -Ct where C is in m/s³. We'll solve for the value of C and the distance traveled in 8.00 s.

Part (a): Finding the value of C

To determine the value of C, consider the velocity function:

v(t) = v₀x + ∫ax dt = v₀x + ∫-Ct dt

Integrating the acceleration to get the velocity:

v(t) = 20.0 m/s - (C/2)t²

Given that the object stops at t = 8.00 s, set v(8.00) = 0:

0 = 20.0 m/s - (C/2)(8.00 s)²

Solving for C:

20.0 m/s = C × 32 s²

C = 40/32 = 1.25 m/s³

Part (b): Distance traveled in 8.00 s

The displacement function x(t) can be found by integrating the velocity function:

x(t) = ∫v(t) dt

x(t) = ∫[20.0 m/s - (C/2)t²] dt

x(t) = 20.0t - (C/6)t³

Using C = 1.25 m/s³ and t = 8.00 s:

x(8.00) = 20.0(8.00) - (1.25/6)(8.00)³

x(8.00) = 160.0 - (1.25/6)(512)

x(8.00) = 160.0 - 106.67

x(8.00) = 53.33 m

Answer:

In the last second, the object traveled 45.6 m.

Explanation:

Hi there!

The equation of the height of the object at a time t is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h =height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s²).

First, let's find how much time it takes the object to reach the ground. For that, we have to find the value of "t" for which h = 0.

h = h0 + v0 · t + 1/2 · g · t²

Since the object is dropped and not thrown (v0 = 0).

h = h0 + 1/2 · g · t²

0 m  = 128 m - 1/2 · 9.8 m/s² · t²

-128 m / -4.9 m/s² = t²

t = 5.1 s

Now, let's find the height of the object 1 s before reaching the ground (at t = 4.1 s):

h = h0 + v0 · t + 1/2 · g · t²

h = 128 m - 1/2 · 9.8 m/s² · (4.1 s)²

h = 45.6 m

Then, in the last second (from 4.1 s to 5.1 s) the object traveled 45.6 m

Answer

Given,

Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.

refractive index of water, n_a = 1.33

refractive index of glass, n_g = 1.52

When the light enters the water, water will act as a lens and when we see the object from the periscope the object shown is farther than the usual distance.

Answer:

N 204.13

Explanation:

Using equation of motion

v² = u² + 2as

u = 0 is the egg was dropped from rest.

v = 2 × 9.8 × 32 = √627.2 = 25.04 m/s

when the egg hit the foam-rubber, the acceleration can  be calculated with

a = change in velocity / change in time = - 25.04 / 0.0092 = -2721.74 m/s²

a) how much it is compressed

v² = u² + 2as

- u² = 2 (-2717.4) s

- 627.2 / -5443.48 = s

s = 0.1152 m = 11.52 cm

b) average force exerted on the egg = mΔv / Δt = 25.04 × 0.075 / 0.0092 = 204.13

Answer:

Part A : 5.) = 564.704 N*m²/C

Part B:  10.) = 179751.0 V/m

Explanation:

A) Applying Gauss'Law to the straight filament, using a cylindrical gaussian surface with the filament as the central axis of the surface, assuming that the electric field is normal to the surface (which means that no flux exist through the lids of the cylinder) and is constant at any point of the surface (except the lids where is 0), we can find the electric flux, as follows:

[tex]E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)[/tex]

where:

r is the radius of the cylinder = 0.05 m

l is the length of the cylinder = 0.01 m

Qenc, is the net charge on the filament enclosed by the gaussian surface

ε₀ = 8.8542*10⁻¹² C²/N*m²

In order to find the value of Qenc, we need to find first the linear charge density, as follows:

[tex]\lambda = \frac{Q}{L} =\frac{+3e-6C}{6m} = 5e-7 C/m[/tex]

The net charge enclosed by the gaussian surface will be just the product of the linear change density λ times the length of the gaussian surface:

[tex]Qenc = \lambda * l = 5e-7C/m * 0.01 m = 5e-9 C[/tex]

According Gauss ' Law, the net flux through the gaussian surface must be equal to the charge enclosed by the surface, divided by the permittivity of free space (in vacuum or air), as follows:

[tex]Flux = \frac{Qenc}{\epsilon 0} = \frac{5e-9C}{8.8542e-12 C2/N*m2} = 564.704 (N*m2)/C[/tex]

which is the same as the option 5.

B)  Repeating the equation (1) from above:

[tex]E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)[/tex]

we can solve for E, as follows:

[tex]E = \frac{Qenc}{2*\pi*r*l* \epsilon 0} = \frac{5e-9C}{2*\pi*(0.05m)*(0.01m)*8.8542e-12 C2/N*m2} = 179751.0 V/m[/tex]

which is the same as the option 10 of part B.

Answer:

The number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³

Explanation:

[tex]N_v = N*e[^{-\frac{Q_v}{KT}}] = \frac{N_A*\rho _F_e}{A_F_e}e[^-\frac{Q_v}{KT}}][/tex]

where;

N[tex]_A[/tex] is the number of atoms in iron = 6.022 X 10²³ atoms/mol

ρFe is the density of iron = 7.65 g/cm3

AFe is the atomic weight of iron = 55.85 g/mol

Qv is the energy vacancy formation = 1.08 eV/atom

K is Boltzmann constant = 8.62 X 10⁻⁶ k⁻¹

T is the temperature = 850 °C = 1123 k

Substituting these values in the above equation, gives

[tex]N_v = \frac{6.022 X 10^{23}*7.65}{55.85}e[^-\frac{1.08}{8.62 X10^{-5}*1123}}]\\\\N_v = 8.2486X10^{22}*e^{(-11.1567)}\\\\N_v = 8.2486X10^{22}*1.4279 X 10^{-5}\\\\N_v = 1.18 X 10^{18}cm^{-3} = 1.18 X 10^{24}m^{-3}[/tex]

Therefore, the number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³

The number of vacancies will be "1.18 × 10²⁴ m⁻³".

According to the question,

Number of atoms, [tex]N_A[/tex] = 6.022 × 10²³ atoms/mol

Iron's density, ρFe = 7.65 g/cm³

Iron's atomic weight, AFe = 55.85 g/mol

Energy vacancy formation, Qv = 1.08 eV/atom

Boltzmann constant, K = 8.62 × 10⁻⁶ k⁻¹

Temperature, T = 850°C or, 1123 K

We know the formula,

→ [tex]N_v[/tex] = N × e [[tex]-\frac{Qv}{KT}[/tex]]

        = [tex]\frac{N_A\times \rho Fe}{AFe}[/tex] e [[tex]-\frac{Qv}{KT}[/tex]]

By substituting the above values, we get

        = [tex]\frac{6.022\times 10^{23}\times 7.65}{55.85}[/tex] e [[tex]- \frac{1.08}{8.62\times 10^{-5}\times 1123}[/tex]]

        = 8.2486 × 10²² × [tex]e^{(-11.1567)}[/tex]

        = 8.2486 × 10²² × 1.4279 × 10⁻⁵

        = 1.18 × 10¹⁸ cm⁻³ or,

        = 1.18 × 10²⁴ m⁻³

Thus the answer above is correct.

Find out more information about vacancy formation here:

https://brainly.com/question/13622505

The frequency of the loop is 58 Hz and the amplitude of the emf induced in the loop is 7.73 mV and this can be determined by using the given data.

Given :

A stiff wire bent into a semicircle of radius 'a' is rotated with a frequency f in a uniform magnetic field.

According to the data, the angular speed is 58 rev/sec that is, 364.4 rad/sec, the magnetic field is 15 mT that is, 15 [tex]\times[/tex] [tex]10^{-3}[/tex] T, and the radius 'a' is 3 cm that is, 3  [tex]\times[/tex] [tex]10^{-2}[/tex] m.

a) The frequency is 58 rev/sec that is 58 Hz.

b) The amplitude of the emf induced in the loop can be calculated as:

[tex]\rm \epsilon = \dfrac{\omega B \pi a^2}{2}[/tex]

Now, substitute the values of the known terms in the above formula.

[tex]\epsilon = \dfrac{364.4\times 15\times10^{-3}\times \pi \times (3\times 10^{-2})^2}{2}[/tex]

Further, simplify the above expression.

[tex]\rm \epsilon = 0.007728\;V[/tex]

[tex]\rm \epsilon = 7.73\;mV[/tex]

For more information, refer to the link given below:

https://brainly.com/question/4393505

Answer

70.52°

Explanation

The distance between projectile's position and it's starting point at any time is given by the relation

r² = x² + y²

where x = horizontal distance covered and y = vertical distance covered

According to projectile motion the horizontal displacement is given by

x = v(x)t = v cos(θ) t

Also the vertical component is given by

y = v(y) t - 0.5gt² = v sin(θ) t - 0.5gt²

Substituting the x and y values into the r-equation yields,

r² = (v cos(θ) t)² + (v sin(θ) t - 0.5gt²)²

r² = v²(cos²(θ))t² + v²(sin²(θ))t² – (vg sin(θ))t³+ 0.25 g²(t^4)

r² = v²t² (cos²(θ)+ sin²(θ)) – (vg sin(θ))t³ + 0.25 g²(t^4)

r² = v²t² – (vg sin(θ))t³ + 0.25 g²(t^4)

Differentiate r with respect to t

r(dr/dt) = 2v²t - 3vg sin(θ)t² + g²t³

At maximum angle the projectile could have been thrown above the horizontal, dr/dt = 0

2v²t - 3vg sin(θ)t² + g²t³ = 0

Divide through by t

2v² - 3vg sin(θ)t + g²t² = 0

g²t² - 3vg sin(θ)t + 2v² = 0

This can be solved using the general law for quadratic equations

(-b ± √(b² - 4ac))2a

a = g², b = -3vg sin(θ) c = 2v²

t = ((3vg sin(θ)) ± √(9v²g²sin²(θ) - 8g²v²))/2g²

This equation makes sense when the value under the square root is positive, that is, the square root exists.

9v²g²sin²(θ) - 8g²v² > 0

9sin²(θ) - 8 > 0

Meaning sin²(θ) = 8/9

Sin θ = (2√2)/3

θ = 70.52°

QED!!!

Answer:

R =  16.67 m

Explanation:

Given:

- Initial Temperature T_i = 20 C

- Thickness of both strips t = 0.001 m

- Final Temperature T_f = 30 C

- Length of the strip L = 0.1 m

- coefficient of linear expansion for brass a_b = 19*10^-6

- coefficient of linear expansion for steel a_s = 13*10^-6

Find:

What is the radius of curvature of this arc R?

Solution:

- The radius of curvature R in relation to dT and a_b, a_s:

                               R = t / dT*(a_b - a_s)

                               R = 0.001 / (30-20)*(19-13)*10^-6

                               R =  16.67 m

Answer:

So after stretching new resistance will be 0.1823 ohm

Explanation:

We have given initially length of the wire [tex]l_1=150m[/tex]

Radius of the wire [tex]r_1=0.32mm=0.32\times 10^{-3}m[/tex]

Resistance of the wire initially [tex]R_1=90ohm[/tex]

We know that resistance is equal to [tex]R=\frac{\rho l}{A}[/tex] ,here [tex]\rho[/tex] is resistivity, l is length and A is area

From the relation we can say that [tex]\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}[/tex]

Now length of wire become 6.75 m

Volume will be constant

So [tex]A_1l_1=A_2l_2[/tex]

So [tex]\pi \times (0.32)^2\times150=\pi \times r_2^2\times 6.75[/tex]

[tex]r_2=1.508mm[/tex]

So [tex]\frac{90}{R_2}=\frac{150}{6.75}\times \frac{1.508^2}{0.32^2}[/tex]

[tex]R_2=0.1823ohm[/tex]

With the new length and cross-sectional area, we determine the new resistance to be approximately 1822.5 Ohms.

To determine the resistance of the wire after it is stretched, follow these steps:

Calculate the initial volume of the wire using the initial length and cross-sectional area.

The initial length (L1) = 1.50 m

The radius (r1) = 0.32 mm = 0.00032 m

Initial cross-sectional area (A1) = πr1² = π (0.00032 m)² = 3.216 × 10⁻⁷ m²

Initial volume (V) = A1 × L1 = 3.216 × 10⁻⁷ m² × 1.50 m = 4.824 × 10⁻⁷ m³

Since volume remains constant, calculate the new radius after stretching.

The new length (L2) = 6.75 m

Initial volume (V) = New volume (V)

V = A2 × L2; thus, A2 = V / L2 = 4.824 × 10⁻⁷ m³ / 6.75 m = 7.148 × 10⁻⁸ m²

New radius (r2) = √(A2 / π) = √(7.148 × 10⁻⁸ m² / π) ≈ 0.000151 m = 0.151 mm

Calculate the new resistance using the resistivity formula.

Resistance (R) = ρ × L / A

Assuming resistivity (ρ) is the same, R1 / R2 = (L1 / A1) / (L2 / A2)

New resistance (R2) = R1 × (L2 / L1)² = 90 Ω × (6.75 m / 1.50 m)²

R2 = 90 Ω × (4.5)² = 90 Ω × 20.25 ≈ 1822.5 Ω

Therefore, the resistance of the wire after stretching is approximately 1822.5 Ohms.

An average adult of weigh 70 kg has approximately [tex]7.0 \times 10^{27}[/tex].

Explanation:

Estimation of Atoms in Human Body

We generally evaluate the amount of elements, liquid, or solid bones in a human body but, we can relate these counts to the atomic level as well. Talking about the approximate amount of Hydrogen, Oxygen and Carbon that we carry in our body, they all make 99% of it neglecting 1% of trace elements.

On the basis of this calculation, we can say that a human body of weight 70 kg has around [tex]7.0 \times 10^{27}[/tex] atoms that are differentiated as 2/3 of hydrogen atoms, 1/4 of Oxygen and only 1/3 of Carbon atoms in the total count.

Answer:

[tex]v_0 = 3.53~{\rm m/s}[/tex]

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

[tex]x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}[/tex]

For the y-direction gives

[tex]v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t[/tex]

Combining both equation yields the y_component of the final velocity

[tex]v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}[/tex]

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

[tex]\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}[/tex]

Final answer:

The question is about projectile motion in Physics, requiring knowledge of equations of motion, range calculation, and trigonometry to find the release angle for an arrow to hit a target and resolve if it will clear an obstacle.

Explanation:

The subject of this question is Physics, specifically dealing with concepts related to projectile motion and the laws of motion. The grade level would likely be High School where students are typically introduced to these topics in a physics curriculum.

As an example, for the question where an archer shoots an arrow at a 75.0 m distant target with an initial speed of 35.0 m/s, one would need to use the equations of projectile motion to solve for the correct release angle. This involves breaking down the initial velocity into its horizontal and vertical components, using the range equation for projectile motion, and applying trigonometry. To determine whether the arrow will go over or under a branch 3.50 m above the release point, one would need to calculate the maximum height of the arrow's trajectory.

In scenarios such as these, understanding the principles of inertia, conservation of momentum, and the conversion of potential energy into kinetic energy are essential.