How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from the horizontal?

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

[tex]KE_i = KE_f[/tex]

[tex]KE_f = \frac{1}{2} mv^2[/tex]

Here,

m = Mass

V = Velocity

Replacing,

[tex]KE_f = \frac{1}{2} (12000)(11)^2[/tex]

[tex]KE_f = 72600J[/tex]

Therefore the  final kinetic energy of the two car system is 72.6kJ

The final kinetic energy of two car system is 72,600 Joules.

To understand more, check below explanation.

The energy is neither be created nor be destroyed only transfer from one form to another form.

So that,

      Initial kinetic energy = Final kinetic energy

It is given that, mass,m = 1200kg , speed, v = 11m/s.

As we know that,

        Kinetic energy[tex]=\frac{1}{2}*m*v^{2}[/tex]

Substute above values in above formula.

        [tex]K.E=\frac{1}{2}*1200*(11)^{2} \\\\K.E=600*121=72,600J[/tex]    

Hence, the final kinetic energy of two car system is 72,600 Joules.

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Answer:

12J

Explanation:

Kinetic Energy before collision = 1/2mv1^2 = 1/2×2×4^2 = 16J

Velocity after collision (v2) = m1v1/m1+m2 = 2×4/2+2 = 8/4 = 2m/s

Kinetic Energy after collision = 1/2mv2^2 = 1/2×2×2^2 = 4J

Kinetic Energy lost = 16J - 4J = 12J

Answer:

Lost in kinetic energy = 12 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')..................................... Equation 1

Where m = mass of first body, u = initial of the first body, m' = mass of the second body, u' = initial velocity of the second body, V = common velocity.

Making V the subject of the equation,

V = mu+m'u'/(m+m')........................... Equation 2

Where m = 2.0 kg, m' = 2.0 kg, u = 4.0 m/s, u' = 0 m/s ( stationary).

Substitute into equation 2

V = (2×4 + 2×0)/(2+2)

V = 8/4

V = 2 m/s.

Total kinetic energy before collision = 1/2mu² = 1/2(2)(2)² = 16 J.

Total Kinetic energy after collision = 1/2V²(m+m') = 1/2(2²)(4) = 4 J.

Thus

Lost in kinetic energy = 16 - 4

Lost in kinetic energy = 12 J

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec[/tex]

Velocity of the string wave is equal to [tex]v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec[/tex]

Power of wave propagation is equal to [tex]P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt[/tex]

So power of the wave will be equal to 5.464 watt

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

[tex]r=(100\cos2\theta)\ m[/tex]

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

[tex]v=r\omega[/tex]

[tex]\omega=\dfrac{v}{r}[/tex]

Where, v = velocity

r = radius

Put the value into the formula

[tex]\omega=\dfrac{29.0576}{100\cos30}[/tex]

[tex]\omega=0.335\ rad/s[/tex]

Hence, The rate at which radar must rotate is 0.335 rad/s.

Answer:

The net work produced is 30.37 KJ

The efficiency of cycle is 72.3%

Explanation:

For the net work produced, we have the formula:

Work = (Th - Tc)(Sh - Sc)(M)

Where,

Th = higher temperature = 800° C + 273 = 1073 k

Tc = lower temperature = 25° C + 273 = 298 k

Sh = specific entropy at higher temperature

Sc = specific entropy at lower temperature

M = molar mass of air

Using, ideal gas table to find entropy. The table is attached.

therefore,

Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)

Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)

Work = 30.37 KJ

Now, for the efficiency (n), we have a formula:

n = 1 - Tc/Th

n = 1 - (298 k)/(1073 k)

n = 0.723 = 72.3 %

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

[tex]m_1[/tex] = Mass of Earth

[tex]m_2[/tex] = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

[tex]F_o=\dfrac{Gm_1m_2}{r^2}[/tex]

New gravitational force

[tex]F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}[/tex]

Dividing the equations

[tex]\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4[/tex]

The ratio is [tex]\dfrac{F_n}{F_o}=4[/tex]

The new force would be 4 times the old force

If the distance between the Earth and Moon were halved, the force of gravity between them would be quadrupled.

According to Newton's law of universal gravitation, the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, this relationship is expressed as:

F= G*m*M/r^2

If the distance between the Earth and Moon were halved, the force of gravity between them would be quadrupled. This is because the force of gravity is inversely proportional to the square of the distance between two objects. So, if the distance is divided by 2, the new force would be multiplied by (2^2) = 4.

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The flexural strength of the ZrO2 block is 7680 lb/in^2. The flexural modulus of the ZrO2 block is 33546.74 lb/in^2.

To calculate the flexural strength of the ZrO2 block, we need to use the formula:

Flexural strength = (3F * L) / (2 * b * h^2)

where F is the applied force, L is the distance between supports, b is the width of the block, and h is the thickness of the block. Substituting the given values, we have:

Flexural strength = (3 * 400 lb * 8 in.) / (2 * 0.50 in. * (0.25 in.)^2) = 7680 lb/in^2

To calculate the flexural modulus, we can use the formula:

Flexural modulus = (F * L^3) / (4 * b * h^3 * y)

where y is the deflection of the block. Substituting the given values, we have:

Flexural modulus = (400 lb * (4 in.)^3) / (4 * 0.50 in. * (0.25 in.)^3 * 0.037 in.) = 33546.74 lb/in^2

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To solve this problem we will apply the concept of Newton's second law with which we will obtain the strength of the proton. We know the mass and the acceleration is given in the statement. Subsequently said Force by equilibrium can be matched the electrostatic force of Coulomb, defined as the product between the charge and the electric field. Our values are

[tex]m = 1.67*10^{-27}kg[/tex]

[tex]a = 10*10^{11} m/s^2[/tex]

Applying the Newton's second law,

[tex]F = ma[/tex]

[tex]F = (1.67*10^{-27}kg)( 10*10^{11} m/s^2)[/tex]

[tex]F = 1.67*10^{-15}N[/tex]

By the Coulomb's equation for electrostatic Force we have that

[tex]F = qE[/tex]

Remember that the charge of a proton is [tex]1.6*10^{-19}C[/tex]

Replacing we have,

[tex]1.67*10^{-15} = 1.6*10^{-19} E[/tex]

[tex]E = 10437.5 N/C[/tex]

Therefore the magnitude of the electric field at the proton's location is [tex]10437.5 N/C[/tex]

The magnitude of the electric field at the proton's location is 10,437.5 N/C.

The given parameters:

The magnitude of the electric field at the proton's location is calculated as follows;

F = ma

F = qE

qE = ma

[tex]E = \frac{ma}{q} \\\\E = \frac{1.67 \times 10^{-27} \times 10\times 10^{11}}{1.6 \times 10^{-19}} \\\\E = 10,437.5 \ N/C[/tex]

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Answer:

(a). D(t) = (√2gh)t - 1/2gt²

(b). D(t) = tk [(√2gh) + g(tk - 2t) / 2]

Explanation:

we would be using the equation of motion to determine the distance D

from the question;

H-boris = h₀ + V₀ (t - tk) - 1/2g(t -tk)²

H-arabella = h₀ + V₀t - 1/2gt²

(a). the vertical displacement is given;

D(t) = H-arabella - H-boris

D(t) = h₀ + V₀t -1/2gt₂ - (h₀ + V₀(t-tk) -1/2g(t-tk)₂)

D(t) = h₀ + V₀t - 1/2gt₂ - h₀ where 0 ∠ t ∠ tk

this gives D(t) = V₀t -1/2gt²

where V₀ = √2gh

∴ D(t) = (√2gh)t - 1/2gt²

(b). We already know vertical displacement as;

D(t) = H-arabella - H-boris

D(t) = h₀ + V₀t -1/2gt₂ - (h₀ + V₀(t-tk) -1/2g(t-tk)₂)

= V₀tk - 1/2gt² + 1/2g(t -tk)²

  = V₀tk + 1/2gtk² - gttk

  = √2gh tk + 1/2gt² - gttk

this gives D(t) = tk [(√2gh) + g(tk - 2t) / 2]

cheers i hope this helps.

Answer:

Explanation:

Point charge is Placed at the center of a spherical Gaussian surface and according to Gauss law electric flux through Gaussian surface is given by [tex]\frac{1}{\epsilon _0}[/tex] times the charge enclosed

where [tex]\epsilon _0=[/tex]permittivity of free space

So unless we take charge outside of the surface flux will not change for the Gaussian surface

So Statement A,B,C and E are false

Option D is correct because we take the charge outside of the Gaussian surface                    

The electric flux depends on the charge enclosed by the Gaussian surface and the location of the charges. If the volume of the surface changes, but the charge remains inside, the flux remains the same. If the charge is moved off-center or outside the surface, or if a second charge is added outside the surface, the flux will change.

The electric flux depends on the charge enclosed by the Gaussian surface. So, if the sphere is replaced by a cube of the same volume with the charge at the center, the electric flux will not change. However, if the cube is one-tenth the volume of the original sphere, the electric flux will decrease by a factor of 10, since the charge is still at the center.

If the point charge is moved off center, but still inside the original sphere, the electric flux will not change, as long as the charge remains enclosed by the Gaussian surface.

If the point charge is moved to just outside the sphere, the electric flux will increase, as some electric field lines are now passing through the Gaussian surface.

If a second point charge is placed just outside the sphere, the electric flux through the Gaussian surface will increase even more, as the electric field lines from both charges are now passing through the surface.

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Answer:

The coyote crashes 66 m from the base of the cliff.

Explanation:

Hi there!

The equation of the position vector of the Coyote is the following:

r = (x0 + v0 · t + 1/2 · a · t², y0 + 1/2 · g · t²)

Where:

r = postion vector of the Coyote at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

a = horizontal acceleration.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference at the edge of the cliff so that x0 and y0 = 0.

When the Coyote reaches the ground, the vertical component of its position vector (r1 in the figure) will be equal to -29 m. When the vertical component of the position vector is -29 m, the horizontal component will be equal to the horizontal distance traveled by the Coyote (r1x in the figure). So, let's find the time at which the y-component of the position vector is -29 m:

y = y0 + 1/2 · g · t²  (y0 = 0)

-29 m = -1/2 · 9.8 m/s² · t²

t² = -29 m / -4.9 m/s²

t = 2.4 s

Now, let's find the x-component of the vector r1 in the figure:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0)

x = 24 m/s · 2.4 s + 1/2 · 3 m/s² · (2.4 s)²  

x = 66 m

The coyote crashes 66 m from the base of the cliff.

The problem involves determining the distance Wile E. Coyote travels horizontally before he crashes into the ground. It involves the use of motion equations to first calculate the time it takes for the coyote to hit the ground and then the distance it travels horizontally. The calculated distance is roughly 65.4 meters.

In this problem, we are asked to calculate the distance that Wile E. Coyote travels along the ground before he crashes.

To solve this, we need to use the laws of motion. First, since the coyote falls with a constant acceleration due to gravity, we can use the equation of motion to determine the time it takes for him to hit the ground:

29 = 0.5 * g * t2

By plugging in the acceleration due to gravity (g=9.8m/s2), we determine that t is approximately 2.44 seconds.

Next, we have to calculate how far the coyote travels horizontally. He starts with a velocity of 24m/s, but he also accelerates due to the rocket. So the distance he travels is given by:

x = v0t + 0.5*a*t2

Substituting the known values (v0 = 24m/s, a = 3m/s2, t = 2.44s), we find that the coyote travels roughly 65.4 meters along the ground before he crashes into it.

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Answer:

E along –x; F along +x

Explanation:

The electric field at point P due to the negative point charge is directed along the -x axis, and the force experienced by the electron at point P due to this field is also along the -x axis.

The electric field at point P due to the negative point charge at x = 0 is directed along the -x axis. This is because electric field lines always point away from positive charges and toward negative charges. Since the negative charge is located at x = 0, the electric field at point P, which is on the positive x-axis, points in the opposite direction, i.e., along the -x axis.

The force experienced by the electron at point P due to this electric field will be in the same direction as the electric field, i.e., along the -x axis. Like charges repel each other, so the negative point charge will exert a repulsive force on the electron.

The angle of incidence, when there is perfect transmission is the polarization angle. The expression for the polarization is

[tex]\theta = 90-\phi[/tex]

Where

[tex]\theta[/tex] = Polarization angle

[tex]\phi[/tex] = Angle with the vertical axis

We have that the angle is

[tex]\theta = 90-19.5[/tex]

[tex]\theta = 70.5\°[/tex]

The ratio of the intensities depends on the cosine of the polarization angle. The polarization angle found from the wearer’s leaning angle can be used to find the fraction of the reflected ray intensity that will pass through the sunglasses.

Applying the Malus Law we have that

[tex]\frac{I}{I_0} = cos^2 \theta[/tex]

Here,

[tex]I[/tex] = Final intensity

[tex]I_0[/tex] = Initial intensity

Replacing we have,

[tex]\frac{I}{I_0} = cos^2 (70.5)[/tex]

[tex]\frac{I}{I_0} = 0.11[/tex]

Therefore the fraction of the reflected light intensity which passes through the sunglasses is 0.11

By applying Malus's Law, a person leaning at 19.5° while wearing polarized sunglasses with a vertical axis lets approximately 88.9% of the reflected, polarized light pass through.

                  I = Io cos²(θ)

                  I = Io cos²(19.5°)

                  cos(19.5°) ≈ 0.943

                  (0.943)² ≈ 0.889

Therefore, the fraction of the reflected light intensity that will pass through the sunglasses is approximately 0.889 or 88.9%.

Answer:

Distance traveled by projectile in horizontal direction will be 4.95 m  

Explanation:

We have given horizontal velocity of the projectile v = 9 m/sec

We have to find the distance traveled by projectile in 0.550 sec in horizontal direction.

We know that distance is equal to multiplication of speed and time

So distance traveled in horizontal direction will be equal to [tex]d=v\times t=9\times 0.550=4.95m[/tex]

So distance traveled by projectile in horizontal direction will be 4.95 m  

Answer:

4.95 m

Explanation:

Horizontal velocity of projectile, u = 9 m/s

time, t = 0.550 s

The horizontal distance, d = horizontal velocity x time

d = 9 x 0.55

d = 4.95 m

Thus, the horizontal distance traveled is 4.95 m.

Answer: The heart pumps 124.2 billion cm³ of blood in a lifetime

Explanation:

as an adult the pulse rate average must be around 72 beats per minute.

The heart beats about 103,680 times in a day.

There are 365 days in a year

number of heart beat in a year = 365 days x 103,680 = 37,843,200 beats in a year

For every the heart pumps 50cm³ of blood,

Hence,

Amount of blood pump in a year = 50 x 37,843,200 = 1,892,160,000cm³ of blood pumped in a year.

Using the estimated lifespan average an individual is 69 years

So in a life time,

The human heart pumps = 1,892,160,000 x 69 years = 124,200,000,000

If the heart pumps 50cm³ of blood per beat, the heart pumps a total of 130,559,040,000 cm³ (130.6 billion cm³) of blood in a LIFETIME.

The human heart beats approximately 108,000 times per day, amounting to nearly 3 billion beats in a 75-year lifespan and pumps about 2.6 million gallons of blood. To estimate the volume in cubic meters, we use the flow rate of 5 L/min over a 75-year period and convert liters to cubic meters.

Estimating Human Heart Beat and Blood Volume Pumped Over a Lifetime

The vital importance of the heart is evident in its tireless work throughout a person's life. Assuming an average heart rate of 75 beats per minute, we can estimate that a human heart beats about 108,000 times in one day, which amounts to more than 39 million times in one year, and nearly 3 billion times during a 75-year lifespan. When it comes to the volume of blood pumped, with each contraction pumping approximately 70 mL of blood, the heart pumps roughly 5.25 liters of blood per minute. This translates to about 14,000 liters per day, and over a year, the heart would pump approximately 10,000,000 liters, or roughly 2.6 million gallons of blood through an extensive network of vessels.

Answer:

The change of the bar's length is [tex] 3.9\times10^{-4} m[/tex]

Explanation:

The bar length is a function of temperature T above room temperature:

[tex] L(T)=1.0000+2.4\times10^{-5} T[/tex]

So, if we evaluate at T= 16.1 C above room temperature

[tex]L(16.1)=1.0000+2.4\times10^{-5} (16.1)[/tex]

[tex]L=1.00039 m [/tex]

Now we can find the change of the bar length with the difference of L and Lo (the length at room temperature)

[tex]L- L_0=1.00039-1.0000 = 3.9\times10^{-4} m[/tex]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

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

[tex]-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2[/tex]

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

[tex]t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}[/tex]

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

Answer:

all areas of knowledge that wish to understand the physical, chemical and biological process must know electrostatics

Explanation:

Electrostatic interactions, have many rare manifestations in nature, which causes many reasons to study them.

- Lightning is a very striking form of electricity

- The biological processes are governed by currents of inanes and potential differences

- The transfer of nutrients and fertilizers to plants is with ion exchange, electrostatic forces

- all modern electronics is based on electricity

- the electric charge in very dry places, creates high currents that can create fires or kill people

In summary all areas of knowledge that wish to understand the physical, chemical and biological process must know electrostatics

A non-physicist might be interested in electrostatic interactions due to their real-world applications, such as photocopiers and air filters, as well as their significance in biological systems.

A non-physicist might be interested in electrostatic interactions for a couple of reasons. First, electrostatic interactions play a significant role in several real-world applications, such as photocopiers, laser printers, ink-jet printers, and electrostatic air filters. Understanding these interactions can help in understanding how these technologies work and how they can be improved. Second, electrostatic interactions are also important in biological systems, where they affect the interactions between molecules. This knowledge is relevant in fields like medicine and biology.

The aqueous solution with the highest concentration of particles will have the lowest freezing point.

The freezing point of a solution is determined by the concentration of solute particles in the solution. The greater the concentration of particles, the lower the freezing point. In this case, the aqueous solution with the lowest freezing point will have the highest concentration of particles.

Out of the options given, the aqueous solution with the highest concentration of particles is 0.60 m aqueous sucrose. Therefore, aqueous sucrose (0.60 m) will have the lowest freezing point.

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Answer:

The distinction can be understood by their individual definitions given below.

Explanation:

A meteoroid is a small rocky/metallic body that can be found in outer space (space beyond the Earth's atmosphere). Their sizes are much smaller than asteroids (often called planetoids) and even more smaller than that of any planets or their moons. They generally originate from comets, asteroids (fragments of them) and even from planets or moons when there occurs heavy collisions.

A meteor is basically what we know to be "shooting stars". When a meteoroid, asteroid, etc. passes through the Earth's atmosphere, they heat up and begin to glow because of the frictional force experienced due to gas molecules in the atmosphere. But the important thing is that they do not reach the surface of the Earth as they completely burn out long before coming close. If some object does manage to reach the Earth's surface, we then call it a meteorite.

(These definitions are not restricted to the Earth but applies to all panets and moons.)

(Also check the gif provided here: https://en.wikipedia.org/wiki/Meteoroid)

Answer:

Explanation:

Single turn implies that N=1

The rectangle is 6cm (0.06m) wide and 10 cm (0.1m) long.

Current in loop is 3A

Magnetic Field B= 4×10^-3T

To take the maximum torque then, sinθ=1

Then

τ=NiABsinθ

Where N is number of turns, and in this case N is 1

i is current in coil and it this case it is 3Amps.

A is area of of coil

And in this case it is a rectangle

Area of rectangle is Lenght × breadth

A= 0.06×0.1= 0.006m^2

And B is magnetic field given as B=4×10^-3T

And θ is angle between the field and the normal to the coil.

a. Torque parallel to the short side of the loop,

The length of the short side is 6cm=0.06m.

This is actually the maximum possible torque, when the field is in the plane of the loop.

τ=NiABsinθ

τ=1×3×0.006×4×10^-3

τ=0.072 ×10^-3

τ=7.2 ×10^-5Nm

b. Torque parallel to the long side of the loop,

The length of the short side is 10cm=0.1m.

This is actually the maximum possible torque, when the field is in the plane of the loop.

τ=NiABsinθ

τ=3×0.006×4×10^-3

τ=0.072 ×10^-3

τ=7.2 ×10^-5Nm.

c. Torque perpendicular to the plane. When the field is perpendicular to the loop the torque is zero.

If the torque is perpendicular to the plane, we said theta is the angle between the normal and the magnetic field

Then if the torque is perpendicular to the normal, then the angle between the torque and the normal is 0, the sinθ = 0

τ=NiABsinθ

Since sinθ =0

Then,

τ=0Nm

Incomplete question as we have not told to find any quantity so I have chosen  some quantities to find.So complete question is here

A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations.

Find (a) the force constant of the spring

 (b) the frequency of the oscillations

 (c) the maximum speed of the object.

Answer:

(a) [tex]k=100N/m[/tex]

(b) [tex]f=1.126Hz[/tex]

(c) [tex]v_{max}=1.41m/s[/tex]

Explanation:

Given data

Mass of object m=2.00 kg

Horizontal Force F=20.0 N

Distance of the object from equilibrium A=0.2 m

To find

(a) Force constant of the spring k

(b) Frequency F of the oscillations

(c) The Maximum Speed V of the object

Solution

For (a) force constant of the spring k

From Hooke's Law we know that:

[tex]F=kx\\k=F/x\\where\\x=A(amplitude)\\So\\k=F/A\\k=(20N/0.2000m)\\k=100N/m[/tex]

For (b) frequency F of the oscillations

The frequency of the motion for an object in simple harmonic motion is expressed as:

[tex]f=\frac{1}{2\pi } \sqrt{\frac{k}{m} }\\ f=\frac{1}{2\pi } \sqrt{\frac{100N/m}{2.0kg} }\\f=1.126Hz[/tex]

For (c) maximum speed V of the object

For an object in simple harmonic motion the maximum values of the magnitude velocity is given as:

[tex]v_{max}=wA\\ where\\w=\sqrt{\frac{k}{m} }\\ so\\v_{max}=\sqrt{\frac{k}{m} }A\\v_{max}=\sqrt{\frac{100N/m}{2.0kg} }(0.200m)\\v_{max}=1.41m/s[/tex]

Answer:

a) 8.83*10⁵ m/s  b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the  force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

[tex]F = e*E = mp*a[/tex]

Solving for a:

[tex]a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2[/tex]

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

[tex]vf^{2} -vo^{2} = 2*a*x[/tex]

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and  x = 0.03 m, we can find vf as follows:

[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s[/tex]

⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s[/tex]

⇒ vf = 2.80*10⁶ m/s

To find the speed of the proton after traveling a certain distance in a uniform electric field, we can use the equations of motion and the equation for force. By plugging in the known values and solving for acceleration, we can then use the equation for speed to find the desired velocity.

To find the speed of the proton, we can use the equation:

F = ma

where F is the force, m is the mass, and a is the acceleration.

The force on the proton is given by:

F = qE

where q is the charge of the proton and E is the electric field strength.

Plugging in the values, we get:

qE = ma

Solving for acceleration, we have:

a = qE/m

For the first part of the question, with a distance of 3.00 cm, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration, and s is the distance.

Plugging in the known values, we have:

v^2 = 0 + 2 * (qE/m) * s

Simplifying:

v = sqrt(2 * (qE/m) * s)

For the second part of the question, with a distance of 30.0 cm, we can use the same equation:

v^2 = u^2 + 2as

Plugging in the known values:

v^2 = 0 + 2 * (qE/m) * s

Simplifying:

v = sqrt(2 * (qE/m) * s)

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Answer:

a) [tex]\Delta s=5\ m[/tex] is the distance between deer and the vehicle

b) [tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.

Explanation:

Given:

a)

Now the distance travelled after application of the brakes till the vehicle stops:

[tex]v^2=u^2+2a_m.s[/tex]

(assuming that the brakes are applied with maximum acceleration)

where:

[tex]s=[/tex] displacement of the vehicle after braking till it stops

[tex]v=[/tex] final velocity of the vehicle = 0 (stops)

putting the values:

[tex]0^2=20^2-2\times 10\times s[/tex]

[tex]s=20\ m[/tex]

Now before the application of the brakes 0.5 second is taken to react and the vehicle travels during this time as well.

So, distance covered before applying the brakes:

[tex]s'=u.t'[/tex]

[tex]s'=20\times 0.5[/tex]

[tex]s'=10\ m[/tex]

The distance between the deer and the vehicle:

[tex]\Delta s=x-(s+s')[/tex]

[tex]\Delta s=35-(20+10)[/tex]

[tex]\Delta s=5\ m[/tex]

b)

The maximum speed the driver can have with the vehicle and still not hit the deer is given as:

[tex]v^2=u'^2+2. a_m.(x-s')[/tex]

because s' is the distance covered before braking during the reaction time.

[tex]0^2=u'^2-2\times 10\times (35-10)[/tex]

[tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.

Using the equations of motion under constant acceleration, a) the distance between the driver and the deer when the car comes to a stop is 5 m and b) the maximum speed the driver could have and still not hit the deer is approximately 23.45 m/s.

The subject of the question, Problem-Solving Strategy 2.1 Motion with constant acceleration, involves using the equations of motion under constant acceleration. Let's break down the problem into two parts:

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Answer:

[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]

Explanation:

We are given that

Displacement along x- axis =-[tex]70\hat{i}[/tex] m

Displacement along y-axis=[tex]19\hat{j}[/tex]m

Displacement along z-axis=[tex]21\hat{k} m[/tex]

Where [tex]\hat{i},\hat{j},\hat{k}[/tex] are unit vector along x, y and z-axis

We have to find the displacement from start to end in unit vector notation.

Total displacement from start to end=Displacement along x-axis+displacement along y-axis+displacement along z- axis

Total displacement from start to end=[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]

Hence, the displacement of the sign from start to end=[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]

To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire. The distance from the line where the electric field strength is 2000N/C is d = lambda / 2 x 10^6 m.

To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire:

E = k * (lambda / d)

Where E is the electric field strength, k is the Coulomb's constant, lambda is the charge density of the wire, and d is the distance from the wire. In this case, since the line of charge is infinite, the charge density is simply the charge per unit length.

To solve for the distance d, we can rearrange the formula: d = k * (lambda / E)

Plugging in the given values, the distance d is:

d = (9.0 x 10^9 Nm^2/C^2) * (lambda / 2000 N/C)

So, the distance from the line where the electric field strength is 2000 N/C is d = lambda / 2 x 10^6 m.

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Answer: 3.35 m/s

Explanation:

This problem is related to the Conservation of Momentum principle, which states the following:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision".

Of course, the momentum of each object may change after the collision, however, the total momentum of the system does not change.  In addition, according to this law it is established that the initial momentum [tex]p_{o}[/tex] must be equal to the final momentum [tex]p_{f}[/tex]:

[tex]p_{o}=p_{f}[/tex] (1)

Where:

[tex]p_{o}=m_{1}V_{1}+m_{2}V_{2}[/tex] (2)

[tex]p_{f}=m_{1}U_{1}+m_{2}U_{2}[/tex] (3)

Being:

[tex]m_{1}=282 kg[/tex] the mass of the first car

[tex]V_{1}=3.70 m/s[/tex] the initial velocity of the first car

[tex]m_{2}=155 kg[/tex] the mass of the second car

[tex]V_{2}=-1.38 m/s[/tex] the initial velocity of the second car

[tex]U_{1}=1.10 m/s[/tex] the final velocity of the first car

[tex]U_{2}[/tex] the final velocity of the second car

Substituting (2) and (3) in (1):

[tex]m_{1}V_{1}+m_{2}V_{2}=m_{1}U_{1}+m_{2}U_{2}[/tex] (4)

Isolating [tex]U_{2}[/tex]:

[tex]U_{2}=\frac{m_{1}(V_{1}-U_{1})+m_{2}V_{2}}{m_{2}}[/tex] (5)

Solving:

[tex]U_{2}=\frac{282 kg(3.70 m/s-1.10 m/s)+(155 kg)(-1.38 m/s)}{155 kg}[/tex] (6)

Finally:

[tex]U_{2}=3.35 m/s[/tex]

Answer:

3.35

Explanation:

Answer:

[tex] v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s[/tex]

Explanation:

For this case we know that the initial velocity is given by:

[tex] v_i = 7 \frac{m}{s}[/tex]

The final velocity on this case is given by:

[tex] v_f = 13 \frac{m}{s}[/tex]

And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:

[tex] v_f = v_i +at[/tex]

If we solve for the acceleration we got:

[tex] a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}[/tex]

So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:

[tex] v_f = v_ i +a t[/tex]

And if we replace we got:

[tex] v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s[/tex]

Answer:

(a) The time the book is in the air stays the same.

(b) The horizontal distance the book travels doubles.

(c) The speed of the book just before it reaches the floor increases but not doubles.

Explanation:

The following kinematics equations will be used to solve this question:

[tex]v_y = v_{y_0} + a_yt\\y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)[/tex]

(a) Initially, the y-component of the velocity is zero. So, the x-component of the velocity is doubled.

We will use the second equation for both cases:

[tex]0 - y_0 = v_{y_0}t - \frac{1}{2}(g)t^2 = 0 - \frac{1}{2}gt^2\\0 - y_0 = v_{y_0}t_2 -\frac{1}{2}gt_2^2 = 0 - \frac{1}{2}gt_2^2\\t = t_2[/tex]

Since, the initial velocity in the y-direction is zero, and the height of the table is constant. The time it takes from the edge of the table to the floor is the same.

(b) For the horizontal distance the book travels, we should use the second equation again, and keep in mind that the acceleration in the x-direction is zero.

[tex]x - 0 = v_{x_0}t + \frac{1}{2}a_xt^2 = v_0t\\x_2 = 2v_0t_2 = 2v_0t[/tex]

Hence, the horizontal distance doubles.

(c) The vertical velocity does not change, since the initial velocity in the y-direction is zero. The horizontal velocity does not change along the motion, since the acceleration in the x-direction is zero. However, since the initial velocity in the x-direction doubles, the final velocity in the x-direction doubles as well.

[tex]v_x = v_{x_0} + a_xt = v_{x_0}[/tex]

However, this does not mean that final speed of the book will double. Because the speed of the object is calculated as follows:

[tex]v_{\rm final_1} = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + v_y^2}\\v_{\rm final_2} = \sqrt{v_{x_2}^2 + v_y^2} = \sqrt{(2v_0)^2 + v_y^2}[/tex]

As can be seen from above, the final speed increases but not doubles.

Doubling the initial horizontal velocity of a book in projectile motion would result in the same time in the air, double the distance covered, and a higher (but not doubled) speed when it hits the floor.

This question explores the concepts of projectile motion in physics. When we double the initial horizontal velocity of the book from v0 to 2v0, the following changes occur:

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