An infinitely long line charge of uniform linear charge density λ = -3.00 µC/m lies parallel to the y axis at x = -3.00 m. A point charge of 3.00 µC is located at x = 1.00 m, y = 2.00 m. Find the electric field at x = 2.00 m, y = 1.50 m.

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

[tex]Energy=h\times frequency[/tex]

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So,  Formula for energy:

[tex]Energy=h\times \frac {c}{\lambda}[/tex]

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

[tex]2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}[/tex]

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

Wavelength = 736.67 nm

Answer:

Ball A

Solution:

According to the question:

Two balls A and B are kicked at angle [tex]20^{\circ}[/tex] and [tex]75^{\circ}[/tex] respectively.

The component of the velocity that provides the ball with the forward movement is the horizontal component, i.e., cosine of the angle.

Since, cos[tex]20^{\circ}[/tex] is greater than cos[tex]75^{\circ}[/tex] and thus Ball A will have more speed than ball B at the top most point of the trajectory.

Final answer:

Ball A will have a bigger speed at the highest point of its trajectory in comparison to ball B, as it was kicked at a lower angle of 20°, resulting in a larger horizontal component of the initial velocity when compared to ball B kicked at 75°.

Explanation:

The question is asking about the speeds of two balls at the highest point of their trajectories when they are kicked with the same initial velocity but at different angles. To determine the speed at the highest point, we can use the principles of projectile motion. The speed of a projectile at its highest point consists only of the horizontal component of the initial velocity because the vertical component equals zero at that instant.

Since both balls A and B were kicked with the same initial speed, the horizontal component of that speed can be found using the cosine of their respective launch angles. For ball A, this component is cos(20°) multiplied by the initial speed, and for ball B, it is cos(75°) multiplied by the initial speed. Because the cosine of 20° is greater than the cosine of 75°, ball A will have a larger horizontal component of velocity and thus a higher speed at the peak of its trajectory.

Final answer:

When the separation between two 1 kg masses is decreased by 2/3, the mutual gravitational force between them will increase. Conversely, when the separation is increased by a factor of 3, the mutual gravitational force will decrease.

Explanation:

When the separation between two 1 kg masses is decreased by 2/3, the mutual gravitational force between them will increase. As the separation decreases, the gravitational force between the masses becomes stronger. On the other hand, when the separation is increased by a factor of 3, the mutual gravitational force will decrease. As the separation increases, the gravitational force between the masses becomes weaker.

Answer:

F = 22572N

Explanation:

We will first calculate the acceleration needed to achieve landing speed equals zero:

[tex]V_{f}^{2}=V_{o}^{2}+2*a*\Delta Y[/tex]

[tex]0=18^{2}+2*a*(-165)[/tex]   Solving for a:

[tex]a=0.98 m/s^{2}[/tex]

Now that we have the required acceleration, we need to check all forces acting on the craft:

[tex]F_{thrust} - W = m*a[/tex]   where W = 11400N; m = 11400Kg; a = 0.98m/s2

Solving the equation:

[tex]F_{thrust}=22572N[/tex]

The magniture of thrust required to reduce the lunar craft's velocity to zero can be calculated using the principles of kinematics and forces. The change in kinetic energy is calculated and equated with the work done by the retro-rocket's thrust, in order to find the required thrust.

The physics involved in this problem is related to kinematics and forces. To calculate the amount of thrust needed, we first need to find the change in kinetic energy. The initial kinetic energy is 0.5 * 11400 kg * (18 m/s)^2 and the final kinetic energy is zero since the craft is at rest. Therefore, the change in kinetic energy is -0.5 * 11400 kg * (18 m/s)^2.

By the work-energy theorem, the work done by the retro-rocket's thrust is equal to the change in kinetic energy. Since work done is also equal to force (thrust) multiplied by distance, we can equate these two expressions to find the thrust: Thrust * 165 m = -0.5 * 11400 kg * (18 m/s)^2. Solve for thrust to get the magnitude of thrust needed.

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Answer:

- 32.17 fts/s in the imperial system, -9.8 m/s in the SI

Explanation:

We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):

[tex]a = \frac{dv}{dt}[/tex]

So, if we wanna know the change in velocity, we can take the integral:

[tex]v(t_f) - v(t_i) = \int\limits^{t_f}_{t_i} {\frac{dv}{dt} \, dt = \int\limits^{t_f}_{t_i} a \, dt[/tex]

Luckily for us, the acceleration in this problem is constant

[tex]a \ = \ - g[/tex]

the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:

[tex]t_f = t_i + 1 s[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  \int\limits^{t_i + 1 s}_{t_i} a \, dt[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  \int\limits^{t_i + 1 s}_{t_i} (-g ) \, dt[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  [-g t]^{t_i + 1 s}_{t_i} [/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g (t_i+1s) + g t_i [/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g t_i + -g 1s + g t_i[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g 1s [/tex]

taking g in the SI

[tex]g=9.8 \frac{m}{s^2}[/tex]

this is:

[tex]v(t_i + 1 s) - v(t_i) =  - 9.8 \frac{m}{s} [/tex]

or, in imperial units:

[tex]g=32.17 \frac{fts}{s^2}[/tex]

this is:

[tex]v(t_i + 1 s) - v(t_i) =  - 32.17 \frac{fts}{s} [/tex]

Answer:

D. pre schematic

Explanation:

During the _pre schematic____ developmental stage children are drawing their thoughts instead of merely observations.

Pre schematic stage in children is the age at which they drawing, symbols, spatial figures to express themselves. Their becomes more complex and full of lines as they age. They use single base lines, multiple base lines and fold up views.

Answer:

The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]

Explanation:

Given that,

Speed [tex]v=1.30\times10^{6}\ m/s[/tex]

Radius = 0.350 m

We need to calculate the magnetic field

Using formula of magnetic field

[tex]B =\dfrac{mv}{qr}[/tex]

Where, m = mass of electron

v = speed of electron

q = charge of electron

r = radius

Put the value into the formula

[tex]B=\dfrac{9.1\times10^{-31}\times1.30\times10^{6}}{1.6\times10^{-19}\times0.350}[/tex]

[tex]B=2.11\times10^{-5}\ T[/tex]

Hence, The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]

Answer:

volume of air breath = 495.75 mL per min

Explanation:

given data

oxygen required = 2.4 × 10² mL / Min = 240 mL

inhaled air = 20 %

exhaled air = 16 %

to find out

volume of air per breath

solution

we know here that breathes 12 time every minute

and  we consider here air breath = y per min

we know oxygen remain in body = 20 % - 16 % = 4%

and required 12 time air breath

so air required = 240 / 12 = 19.83 mL

so oxygen breath equation will be

y × 4% = 19.83

y = 19.83 / 0.04

y = 495.75

so volume of air breath = 495.75 mL per min

Answer:

A) 594 Hz

B) 512 Hz

Explanation:

Howdy!

This is a typical Doppler effect problem, whose formula is:

[tex]f = \frac{v+v_{r} }{v+v_{s}} f_{0}[/tex]

Where :

[tex]v_{r}[/tex]

is the velocity of the receiver (in our case is equal to zero)

[tex]v_{s}[/tex]

is the velocity of the source (in our case is 92 km/h) which is positive when the source is moving away (second case) and negative otherwise.

[tex]v_{s}[/tex]

Is the velocity of the wave in the medium.

Before we start calculating we need to have the velocity of the source in the same units as the velocity of the waves:

92km/h = 92*(1000)/3600 m/s = 25.5 m/s

A)

When the source is moving towards the receiver the sign of the velocity  is negative, so:

[tex]f = \frac{343}{343-25.5} 550 Hz[/tex]

    f = 594.1 Hz

B) Now the velocity of the source must change sign:

[tex]f = \frac{343}{343+25.5} 550 Hz[/tex]

    f = 511.9 Hz

As a sanity check we know that when the source is moving towards the source the frequency is higher and when the source is moving away from the source the frequency is lower.

A) The frequency heard by a person standing beside the road in front of the car is 595.3 Hz. B) The frequency heard by a person standing beside the road behind the car is 514.5 Hz.

To solve this problem, we will use the Doppler effect formula, which relates the observed frequency to the actual frequency of the source, the speed of sound, and the relative velocity between the source and the observer. The formula for the observed frequency [tex]\( f' \)[/tex] when there is relative motion between the source and the observer is given by:

[tex]\[ f' = \left( \frac{v + v_o}{v + v_s} \right) f \][/tex]

where:

f is the frequency of the source,

v is the speed of sound in the medium,

[tex]v_o[/tex] is the speed of the observer relative to the medium (positive if moving towards the source, negative if moving away),

[tex]v_s[/tex] is the speed of the source relative to the medium (positive if moving away from the observer, negative if moving towards).

A) For the person standing beside the road in front of the car:

The observer is stationary, so [tex]\( v_o = 0 \)[/tex].

The car is moving towards the observer, so [tex]\( v_s = -92 \text{ km/h} \)[/tex].

We need to convert the speed of the car from km/h to m/s to match the units of the speed of sound:

[tex]\[ 92 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 25.56 \text{ m/s} \][/tex]

Now we can plug the values into the Doppler effect formula:

[tex]\[ f' = \left( \frac{343 \text{ m/s} + 0}{343 \text{ m/s} - 25.56 \text{ m/s}} \right) \times 550 \text{ Hz} \][/tex]

[tex]\[ f' = \left( \frac{343}{317.44} \right) \times 550 \][/tex]

[tex]\[ f' \approx 595.3 \text{ Hz} \][/tex]

B) For the person standing beside the road behind the car:

The observer is again stationary, so [tex]\( v_o = 0 \)[/tex].

The car is moving away from the observer, so [tex]\( v_s = 92 \text{ km/h} \)[/tex] or [tex]\( 25.56 \text{ m/s} \)[/tex].

Using the Doppler effect formula:

[tex]\[ f' = \left( \frac{343 \text{ m/s} + 0}{343 \text{ m/s} + 25.56 \text{ m/s}} \right) \times 550 \text{ Hz} \][/tex]

[tex]\[ f' = \left( \frac{343}{368.56} \right) \times 550 \][/tex]

[tex]\[ f' \approx 514.5 \text{ Hz} \][/tex]

Answer:

a. a = [tex]6.41 m/s^2[/tex]

b. T = -0.81 N

Explanation:

Given,

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

[tex]w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]

From the f.b.d. of the heavier block,

[tex]w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)[/tex]

From eqn (1) and (2), we get,

[tex]w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\[/tex]

[tex]\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\[/tex]

[tex]\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.[/tex]

part (b)

From the eqn (2), we get,[tex]T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N[/tex]

Answer:

(a) 2.793 m/s^2

(b) 0.335 N

Explanation:

Let a be the acceleration in the system and T be the tension in the string.

W1 = 3 N

W2 = 7 N

m1 = 3 /1 0 = 0.3 kg

m2 = 7 / 10 = 0.7 kg

θ = 30°

μ1 = 0.13, μ2 = 0.31

Let f1 be the friction force acting on block 1 and f2 be the friction force acting on block 2.

By the laws of friction

f1 = μ1 x N1

Where, N1 be the normal reaction acting on block 1.

So, f1 = 0.13 x W1 Cosθ = 0.13 x 3 x cos 30 = 0.337 N

By the laws of friction

f2 = μ2 x N2

Where, N2 be the normal reaction acting on block 2.

So, f2 = 0.31 x W2 Cosθ = 0.31 x 7 x cos 30 = 1.88 N

Apply Newton's second law for both the blocks

W1 Sin30 - T - f1 = m1 a

3 Sin 30 - T - 0.337 = 0.3 x a

1.163 - T = 0.3 a ..... (1)

W2 Sin30 + T - f2 = m2 a

7 Sin30 + T - 1.88 = 0.7 x a

1.62 + T = 0.7 a ..... (2)

By solving equation (1) and (2) we get

a = 2.793 m/s^2

(b) Put the value of a in equation (2), we get

1.62 + T = 0.7 x 2.793

T = 0.335 N

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

[tex]V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s[/tex]

Part B. Using Kinematics expression for distance, time and initial velocity

[tex]y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s[/tex]

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

The speed of the rock just before it hits the street is approximately 20.3 m/s, and the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Part A: To determine the speed of the rock just before it hits the street, we can use the concept of conservation of energy. When the rock is at the top of its trajectory, its potential energy is at its maximum and its kinetic energy is at its minimum. When the rock is just before hitting the street, its potential energy is at its minimum (zero) and its kinetic energy is at its maximum. Using the equations for potential energy and kinetic energy, we can calculate the speed of the rock:

Initial potential energy = final potential energy + final kinetic energy

mgh = 1/2m*v^2

where m is the mass of the rock, g is the acceleration due to gravity, h is the height of the building, and v is the final velocity of the rock.

Since the mass of the rock cancels out, we have:

gh = 1/2v^2

Plugging in the values, g = 9.8 m/s^2 and h = 21.0 m, we can solve for v:

v = sqrt(2gh)

v = sqrt(2*9.8*21.0)

v = sqrt(411.6)

v = 20.3 m/s

Therefore, the speed of the rock just before it hits the street is approximately 20.3 m/s.

Part B: To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation for time of flight of a vertically thrown object:

t = 2v/g

where t is the time of flight, v is the initial upward velocity of the rock, and g is the acceleration due to gravity.

Plugging in the values, v = 27.0 m/s and g = 9.8 m/s^2, we can solve for t:

t = 2*27.0/9.8

t = 5.5 s

Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Answer:

altitude of satellite from earth surface is 7140 km

Explanation:

given data:

mass of satellite = 5000kg

altitude of satellite from earth surface can be calculated by using given relation

[tex]p^2 = \frac{(4* \pi^2)(a^3)}{(G*M)}[/tex]

where,

p -  satellite's period in seconds,

a - satellite's altitude in meters,

G - gravitational constant (6.67 x 10^-11 m³/kgs²) and

M -  mass of Earth 5.98 x 10^24 kg

solving for altitude we have

[tex]a^3 =\fracp{(G*M*p^2)}{(4*\pi^2)}[/tex]

[tex]a^3 = 3.637 * 10^{20}[/tex]

a = 7,138,000 meters = 7,140 km

altitude of satellite from earth surface is 7140 km

Answer:

Metal is more dense than water.

Explanation:

As we know, the molecules of the metal are tightly closed as compared to that of water and the density of a material is defined as the mass of the material per unit volume.

In a certain volume of metal, there are more numbers of molecules than that of water in the same amount of volume, therefore the density of metal is greater than that of water.

Also, according to Archimedes' principle, if there is an object in a fluid, then the buoyant force on that object is equal to the weight of the fluid that it displaces.

When the density of the object is larger than that of the fluid then it overcomes that buoyant force and sinks.

Thus, an object sinks in a fluid if its density is larger than that of the fluid and floats otherwise.

Since, the metal sink in water, it means Metal is more dense than water.

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

[tex]v^2=u^2+2as[/tex]

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

[tex]0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s[/tex]

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

[tex]v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s[/tex]

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

[tex]v=u+at[/tex]

where symbols have the usual meaning

Applying the given values we get

[tex]t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds[/tex]

Answer:

16 % and 9 %

Explanation:

Accurate reading = 10 N

reading of first balance = 8.4 N

reading of second balance = 9.1 N

Percentage difference in first reading

= [tex]\frac{True reading - reading of first balance}{True reading}\times 100 %[/tex]

= [tex]\frac{10-8.4}{10}\times 100 %[/tex] = 16 %

Percentage difference in second reading

= [tex]\frac{True reading - reading of second balance}{True reading}\times 100 %[/tex]

= [tex]\frac{10-9.1}{10}\times 100 %[/tex] = 9 %

Answer:

56 m/s

Explanation:

The electirc force applied on the particle by the field will be

F = q * E

F = 3*10^-3 * 80 = 0.24 N

This force will cause an acceleration:

F = m * a

a = F/m

a = 0.24 / 0.02 = 12 m/s^2

The equation for speed under constant acceleration is:

V(t) = V0 + a*t

V(3) = 20 + 12 * t = 56 m/s

The final speed will be  56 m/s.

Answer:

D. measurement and geometry

Explanation:

Building helps children develop their _ measurement and geometry__ skills. The measurement and geometry skill of the children by observing buildings.

Geometry is a method or tool for understanding the relations among shapes and spatial properties. Children can develop spatial reasoning and can visualize shapes in different positions (orientation) when observing a building.

Answer:

c)[tex]F_{net} = 0.640 kN[/tex]

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

[tex]F_{net} = ma[/tex]

here we know

m = 80 kg

for circular motion acceleration is given as

[tex]a_c = \frac{v^2}{R}[/tex]

[tex]a_c = \frac{40^2}{200} = 8 m/s^2[/tex]

now we have

[tex]F_{net} = 80 \times 8[/tex]

[tex]F_{net} = 640 N[/tex]

[tex]F_{net} = 0.640 kN[/tex]

Answer:

Speed of the wave, v = 2.4 m/s

Explanation:

Given that,

Frequency of vibrations, f = 4 Hz

Wavelength of the transverse wave, [tex]\lambda=60\ cm=0.6\ m[/tex]

We need to find the speed of the waves along the wire. Let it is equal to v. So, the relationship is given by :

[tex]v=f\times \lambda[/tex]

[tex]v=4\times 0.6[/tex]

v = 2.4 m/s

So, the speed of waves along the wire is 2.4 m/s. Hence, this is the required solution.

Final answer:

The speed of the transverse wave on the wire, given a frequency of 4.00 Hz and a wavelength of 60 cm, is calculated to be 2.40 meters per second (m/s) using the formula Speed = Frequency × Wavelength.

Explanation:

To determine the speed of a wave, you can use the formula:

Speed = Frequency × Wavelength

In the given scenario, the frequency is 4.00 Hz, and the wavelength is 60 cm, which is equivalent to 0.60 meters (since 1 meter = 100 cm). Thus, the speed of the wave on the wire can be calculated as follows:

Speed = 4.00 Hz × 0.60 m

Speed = 2.40 m/s

Therefore, the speed of the transverse wave along the wire is 2.40 meters per second (m/s).

Answer:

Explanation:

given,

mass of the object = 280 g = 0.28 kg

time period = 0.270 s

total energy of the system  = 4.75 J

                 [tex]\dfrac{1}{2}\ m\ V^2 = 4.75 J[/tex]

maximum speed of the object V =     [tex]\sqrt{ \dfrac{2 \times 4.75}{0.28} }[/tex]

                                              V= 5.82 m / s

(b)   force constant of the spring K = m ω²

where ω = angular frequency = 2π / T

          T= time period = 0.25 s

ω = 25.13 rad / s

K = 0.28 × 25.13²

K = 176.824 N / m

(c). Amplitude of motion A =    [tex]\dfrac{V}{\omega}[/tex]

                                         =     [tex]\dfrac{5.82}{25.13}[/tex]

A = 0.232 m

Answer:

Precise and Reproducible

Explanation:

For a system or a process to be useful a measurement standard for any physical quantity, it must be:

Answer:

precise

reproducible

Explanation:

Answer:

Explanation:

Given

Cheetah  speed=21.8 m/s in 2.55 sec

i.e. its [tex]a=\frac{21.8}{2.55}=8.55 m/s^2[/tex]

Cheetah top speed=28.1 m/s

And 1 m/s is equal to 2.23694 mph

therefore 28.1 m/s is [tex]2.236\times 28.1=62.858 mph[/tex]

Time taken to reach top speed

v=u+at

[tex]28.1=0+8.55\times t[/tex]

[tex]t=\frac{28.1}{8.55}=3.286 s [/tex]

Distance traveled during this time

[tex]v^2-u^2=2as[/tex]

[tex]28.1^2=2\times 8.55\times s[/tex]

[tex]s=\frac{789.61}{2\times 8.55}=46.176 m[/tex]

If cheetah sees a rabbit 122 m away

time taken to reach rabbit

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

[tex]122=0+\frac{1}{2}\times 8.55\times t^2[/tex]

[tex]t^2=\frac{244}{8.55}[/tex]

[tex]t=\sqrt{28.53}=5.34 s[/tex]

Answer:

The frequency of the emitted EM wave by He-Ne laser is [tex]4.74\times 10^{14} Hz[/tex]

Given:

Power emitted by He-Ne laser, P = 1.5 mW = [tex]1.5\times 10^{- 3}[/tex]

Wavelength, [tex]\lambda = 632.8 nm = 632.8\times 10^{- 9} m[/tex]

Solution:

Now, to calculate the frequency, [tex]\vartheta[/tex] of the EM wave emitted:

Energy associated with 1 photon = Power of one photon per sec = [tex]\frac{hc}{\lambda}[/tex]

Therefore, power associated with 'N' no. of photons, P = [tex]N\frac{hc}{\lambda}[/tex]

where

[tex]h = 6.626\times 10^{-34} m^{2}kg/s[/tex] = Planck's constant

Now,

[tex]1.5\times 10^{- 3} = N\frac{6.626\times 10^{-34}\times 3\times 10^{8})}{632.8\times 10^{- 9}}[/tex]

N = [tex]4.77\times 10^{15}[/tex]

Also, we know that:

[tex]c = \vartheta \lambda[/tex]

Thus

[tex]\vartheta = \frac{c}{\lambda} = \frac{3\times 10^{8}}{632.8\times 10^{- 9}}[/tex]

[tex]\vartheta = 4.74\times 10^{14} Hz[/tex]

Answer:

18.21 m

41.1 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Part A

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -8.8}\\\Rightarrow s=18.21\ m[/tex]

On a dry day she would have to start braking 18.21 m away from the intersection

Part B

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -3.9}\\\Rightarrow s=41.1\ m[/tex]

On a wet day she would have to start braking 41.1 m away from the intersection

Answer:

The car is traveling at [tex]8.1366\frac{m}{s}[/tex]

Explanation:

The known variables are the following:

[tex]V_{0} = 56 \frac{km}{h} = 15.56 \frac{m}{s}\\ D=25m\\ t=2.11s\\ V_{f}=?[/tex]

First, from the equation of motion we find the deceleration:

[tex]D=V_{0}*t+\frac{1}{2} a*t^{2} \\ a=\frac{2(D-V_{0})}{t^{2} } \\ a=3.5182\frac{m}{s^2}[/tex]

Then, with the equation for the speed:

[tex]V_{f}=V_{o}+a*t\\ V_{f}=8.1366\frac{m}{s}[/tex]

The speed of the car at impact is approximately 25.72 m/s.

To find the speed of the car at impact, we first need to calculate the deceleration. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

Given that the initial velocity (u) is 56.0 km/h, the distance (s) is 25.0 m, and the time taken (t) is 2.11 s, we can rearrange the equation to solve for a: a = (v - u) / t.

Plugging in the values, we have a = (0 - 56.0 km/h) / 2.11 s = -26.5 m/s². Since the car is decelerating, the acceleration is negative.

Now, to find the final velocity (v) at impact, we can use the equation v² = u² + 2as. Rearranging the equation, we have v² = u² + 2(-26.5 m/s²)(25.0 m) = u² - 2(26.5 m/s²)(25.0 m).

Plugging in the values, we have v² = (56.0 km/h)² - 2(26.5 m/s²)(25.0 m).

Converting the initial velocity (u) to m/s, we get u = 56.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 15.6 m/s.

Substituting the values, we have v² = (15.6 m/s)² - 2(26.5 m/s²)(25.0 m) = 0.16 m²/s² - 26.5 m²/s² * 25.0 m = 0.16 m²/s² - 662.5 m²/s² = -662.34 m²/s².

Taking the square root of both sides, we get v ≈ √(-662.34 m²/s²) ≈ -25.72 m/s.

Since speed is a positive quantity, the speed of the car at impact is approximately 25.72 m/s.

Answer:

a) yield strength

   [tex]\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa[/tex]

b) modulus of elasticity

strain calculation

[tex]\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046[/tex]

strain for offset yield point

[tex]\varepsilon_{new} = \varepsilon_0 -0.002[/tex]

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 [tex]E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}[/tex]

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

 [tex]\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa[/tex]

d) percentage elongation

[tex]\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%[/tex]

e) percentage of area reduction

[tex]\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%[/tex]                            

The mechanical properties of the tensile test specimen, such as yield strength, modulus of elasticity, tensile strength, percent elongation, and percent reduction in area, are determined by performing calculations based on the given dimensions, yield force, maximum load, and deformations observed during the test.

To determine the mechanical properties of a tensile test specimen, the following calculations are performed:

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

[tex]a=\dfrac{v-u}{t}[/tex].............(1)

Now, the second equation of motion is :

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

Put the value of a in above equation as :

[tex]s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2[/tex]

[tex]s=\dfrac{t(u+v)}{2}[/tex]

[tex]u=\dfrac{2s}{t}-v[/tex]

[tex]u=\dfrac{2\times 47}{8.6}-2.3[/tex]

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,      

velocity of the observer = 17 m/s

speed of the sound = 343 m/s    

velocity of the source = 0 m/s    

frequency emitted from the source  = 2550 Hz              

[tex]f = f_0(\dfrac{v-v_0}{v-v_s})[/tex]              

[tex]f = 2550\times (\dfrac{343+17}{343-0})[/tex]

velocity of observer is negative as it is approaching the source.                   f = 2676.38 Hz ≈ 2677 Hz                    

hence, the frequency heard by the observer is equal to 2677 Hz

Answer:

Explanation:

We need to supply 5 Nm to the bolt.

If we have a force of 25 N in the direction (2*i - 3*j), the torque is related to the length of the wrench.

T = F * d

d = T / F

d = 5 / 25 = 0.2 m

Assuming the wrench is perpendicular to the force.

The vector of the force is:

[tex]F = 25 N \frac{2i - 3j}{\sqrt{2^2 + (-3)^2}} = 13.9i - 20.8[/tex]

This is a countr-clockwise direction. If the bolt is right handed (most are) we are tightening it.

Answer:

The car travels 800m

Explanation:

We will need two formulas to solve this question:

[tex]V_{f}=V_{0}+at\\X=V_{0}t+\frac{at^{2}}{2}[/tex]

We can obtain the the time by using the final velocity formula and replacing with known values:

[tex]V_{f}=V_{0}+at\\\\0\frac{m}{s}=80\frac{m}{s}-4\frac{m}{s^{2}}t\\\\4\frac{m}{s^{2}}t=80\frac{m}{s}\\t=20 s[/tex]

Since we know the time, the question will be answered by inputing all the variables needed in the distance formula:

[tex]X=V_{0}t+\frac{at^{2}}{2}\\\\X=80\frac{m}{s}(20s)-\frac{4(20s)^{2}}{2}=800m[/tex]

Final answer:

The race car will travel 800 meters before coming to a stop after deploying a drag parachute causing a deceleration of 4 m/s^2 from an initial speed of 80 m/s.

Explanation:

To calculate how far the race car will travel before it stops after deploying a drag parachute, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled. Considering that the deceleration is constant, the kinematic equation we can use is v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

Since the car comes to a stop, v = 0 m/s. The initial speed is u = 80 m/s and the deceleration provided by the drag parachute is a = -4 m/s2 (negative because it's deceleration). Plugging these values into the equation gives us:

0 = (80 m/s)2 + 2(-4 m/s2)s

Solving for s, we find:

s = (80 m/s)2 / (2 * 4 m/s2)

s = 6400 m2/s2 / 8 m/s2

s = 800 meters

Therefore, the race car will travel 800 meters before coming to a stop after the drag parachute is deployed.