A solenoid with n1 = 1200 turns/m and a current I1 = 2.5 A is filled with a paramagnetic material at a temperature T1 = 320 K. A second solenoid with n2 = 1000 turns/m and a current I2= 0.85 A is filled with the same paramagnetic material at a different temperature T2. The magnetizations are the same in both cases. What is the value of T2?

Answer:

Explanation:

It is given that,

Mass of lump, m₁ = 0.05 kg

Initial speed of lump, u₁ = 12 m/s

Mass of the cart, m₂ = 0.15 kg

Initial speed of the cart, u₂ = 0

The lump of clay sticks to the cart as it is a case of inelastic collision. Let v is the speed of the cart and the clay after the collision. As the momentum is conserved in inelastic collision. So,

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

[tex]v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

[tex]v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}[/tex]

v = 3 m/s

So, the speed of the cart and the clay after the collision is 3 m/s. Hence, this is the required solution.

The speed of the cart and clay after the collision is 3 m/s.

The speed of the cart and clay after the collision is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

0.05(12) + 0.15(0) = v(0.05 + 0.15)

0.6 = 0.2v

v = 0.6/0.2

v = 3 m/s

Thus, the speed of the cart and clay after the collision is 3 m/s.

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The Linear speed of the sphere is mathematically given as

v = 4.1 m/s

Question Parameter(s):

A solid sphere of mass 4.0 kg and radius of 0.12 m starts from rest at the top of a ramp inclined 15°

Generally, the equation for the conservation of energy   is mathematically given as

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2[/tex]

Therefore

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})\\\\mgh = \frac{7}{10}mv^2[/tex]

In conclusion, the Speed is

[tex]v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}[/tex]

v = 4.1 m/s

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The linear speed of the sphere when it reaches the bottom of the ramp is approximately[tex]\( 4.85 \, \text{m/s} \).[/tex]

At the top of the ramp, the potential energy U  of the sphere is given by:

[tex]\[ U = mgh \][/tex]

where m  is the mass of the sphere, g is the acceleration due to gravity (approximately[tex]\( 9.8 \, \text{m/s}^2[/tex], and  h is the height of the ramp.

At the bottom of the ramp, the kinetic energy K of the sphere is given by:

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

where  v  is the linear speed of the sphere.

Since energy is conserved, we have:

U = K

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

We can solve for  v  by canceling the mass m from both sides and multiplying through by 2:

[tex]\[ 2gh = v^2 \][/tex]

Taking the square root of both sides gives us the linear speed  v :

[tex]\[ v = \sqrt{2gh} \][/tex]

Given that the height  h  is 1.2 m and the acceleration due to gravity  g  is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex], we can plug in these values:

[tex]\[ v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.2 \, \text{m}} \][/tex]

[tex]\[ v = \sqrt{23.52 \, \text{m}^2/\text{s}^2} \][/tex]

[tex]\[ v \ =4.85 \, \text{m/s} \][/tex]

Answer:

Part a)

a = 531.7 m/s/s

Part b)

a = 54.25 g

Explanation:

Part a)

Initial speed of the car is given as

[tex]v = 105 km/h[/tex]

now we have

[tex]v = 29.2 m/s[/tex]

now we know that it stops in 0.80 m

now by kinematics we have

[tex]a = \frac{v_f^2 - v_i^2}{2d}[/tex]

so we will have

[tex]a = \frac{0 - 29.2^2}{2(0.80)}[/tex]

[tex]a = 531.7 m/s^2[/tex]

Part b)

in terms of g this is equal to

[tex]a = \frac{531.7}{9.80}[/tex]

[tex]a = 54.25 g[/tex]

Final answer:

The magnitude of the average acceleration of the driver during the collision is approximately -532.09 [tex]m/s^2[/tex], which is about 54.29 g's when expressed in terms of the acceleration due to gravity.

Explanation:

To calculate the magnitude of the average acceleration of the driver during the collision, we can use the following kinematic equation that relates velocity, acceleration, and distance:

[tex]v^2 = u^2 + 2a * s[/tex]

Where:
v is the final velocity (0 m/s, since the driver comes to a stop)

u is the initial velocity (105 km/h, which needs to be converted to m/s)

a is the acceleration (the quantity we want to find)

s is the stopping distance (0.80 m)

First, convert the velocity from km/h to m/s by multiplying by (1000 m/1 km)*(1 h/3600 s) to get approximately 29.17 m/s. Now we can solve for 'a' as follows:

[tex](0)^2 = (29.17 m/s)^2 + 2 * a * (0.80 m)-29.17^2 = 2 * a * 0.80a = -(29.17)^2 / (2 * 0.80)a = -532.09[/tex]

We find that the magnitude of the average acceleration is approximately [tex]-532.09 m/s^2[/tex]. To express this in terms of 'g's, we divide by the acceleration due to gravity [tex](9.80 m/s^2)[/tex]:

[tex]a_g = -532.09 / 9.80a_g =54.29 g's[/tex]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, [tex]m=\dfrac{F}{a}[/tex]

[tex]m=\dfrac{70\ N}{7.5\ m/s^2}[/tex]

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

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

[tex]W=9.34\ kg\times 1.62\ m/s^2[/tex]

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

The mass of the block is approximately 9.33 kg, and when you multiply that by the acceleration due to gravity on the moon (1.62 m/s^2), you get a weight of approximately 15 N. Therefore, the closest answer is 15 N.

To solve this problem, we need to find the mass of the block first. We know on earth, Force (F) = mass (m) * acceleration (a). Given that the force is 70N, and the acceleration is 7.5 m/s, we can solve for m. So, m = F/a = 70N / 7.5 m/s = 9.33 kg (approximately).

Now, let's figure out the weight of the same block on the moon. Weight is calculated as mass times the acceleration due to gravity (Weight = m*g). On the moon, the acceleration due to gravity is 1.62 m/s^2, so Weight = 9.33 kg * 1.62 m/s^2 = 15.1 N (approximately).

So, the closest answer will be 15 N.

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Answer:

[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]

Explanation:

As we know by the theory of expansion the change in the volume of the object is directly proportional to change in temperature and initial volume.

So here we can say

[tex]\Delta V = V_0\gamma \Delta T[/tex]

here

[tex]\gamma[/tex] = coefficient of volume expansion

so we have

[tex]\gamma = \frac{\Delta V}{V_0 \Delta T}[/tex]

now plug in all values

[tex]\gamma = \frac{2.25 cm^3}{(541 cm^3)(273 - 133)}[/tex]

[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]

Answer:

15435 J

Explanation:

Latent heat of fusion, Lf = 334 J/g

Specific heat of ice, ci = 2.1 J / g C

Latent heat of vaporisation, Lv = 2230 J/g

Specific heat of water, cw = 4.18 J / g C

mass, m = 50 g, T = - 5 degree C

There are following steps

(i) - 5 degree C ice converts into 0 degree C ice

H1 = m x ci x ΔT = 50 x 2.1 x 5 = 525 J

(ii) 0 degree C ice converts into 0 degree C water

H2 = m x Lf = 5 x 334 = 1670 J

(iii) 0 degree C water converts into 100 degree C water

H3 = m x cw x ΔT = 5 x 4.18 x 100 = 2090 J

(iv) 100 degree C water converts into 100 degree C steam

H4 = m x Lv = 5 x 2230 = 11150 J

Total heat required

H = H1 + H2 + H3 + H4

H = 525 + 1670 + 2090 + 11150 = 15435 J

Answer:

The magnetic flux density is [tex]2.11\times10^{-6}\ T[/tex]

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

[tex]B =\dfrac{\mu_{0}I}{2r}[/tex]

Where,

r = radius

I = current

Put the value into the formula

[tex]B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}[/tex]

[tex]B=2.11\times10^{-6}\ T[/tex]

Hence, The magnetic flux density is [tex]2.11\times10^{-6}\ T[/tex]

Answer:

(a) 3.13 m/s

(b) 9.9 m/s

(c) 7.73 m/s

Explanation:

u = 0 m/s, g = 9.8 m/s^2

Let v be the velocity of ball as it hit the ground.

(a) h = 0.5 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 0.5

v^2 = 9.8

v = 3.13 m/s

(b) h = 5 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 5

v^2 = 98

v = 9.9 m/s

(c) h = 10 feet = 3.048 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 3.048

v^2 = 59.74

v = 7.73 m/s

Answer:

[tex]I = 1.6 kg m^2[/tex]

Explanation:

Moment of inertia of disc is given as

[tex]I = \frac{1}{2}mR^2[/tex]

now we have

m = 20 kg

R = 20.0 cm = 0.20 m

now we have

[tex]I_{disc} = \frac{1}{2}(20 kg)(0.20 m)^2[/tex]

[tex]I_{disc} = 0.4 kg m^2[/tex]

Now the additional mass of 20 kg is placed on its rim so it will behave as a ring so moment of inertia of that part of the disc is

[tex]I = mR^2[/tex]

m = 20 kg

R = 20 cm = 0.20 m

[tex]I_{ring} = 20(0.20^2)[/tex]

[tex]I_{ring} = 0.8 kg m^2[/tex]

Now four point masses each of the mass of one fourth of mass of disc is placed at four positions so moment of inertia of these four masses is given as

[tex]I_{mass} = 4( m'r^2)[/tex]

here we have

[tex]m' = \frac{m}{4}[/tex]

[tex]I_{mass} = 4(\frac{m}{4})(0.10^2 + 0.10^2)[/tex]

[tex]I_{mass} = 20(0.02) = 0.40 kg m^2[/tex]

Now total moment of inertia of the system is given as

[tex]I = I_{disc} + I_{ring} + I_{mass}[/tex]

[tex]I = 0.4 + 0.8 + 0.4 = 1.6 kg m^2[/tex]

Answer:

The magnitude of the electric flux is [tex]3.53\ N-m^2/C[/tex]

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area [tex]A= 1.65 m^2[/tex]

We need to calculate the flux

Using formula of the magnetic flux

[tex]\phi=E\cdot A[/tex]

[tex]\phi = EA\cos\theta[/tex]

Where,

A = area

E = electric field

Put the value into the formula

[tex]\phi=2.35\times1.65\times\cos 25^{\circ}[/tex]

[tex]\phi=2.35\times1.65\times0.91[/tex]

[tex]\phi=3.53\ N-m^2/C[/tex]

Hence, The magnitude of the electric flux is [tex]3.53\ N-m^2/C[/tex]

Answer:

The shear deformation is [tex]\Delta x=3.34\times10^{-6}\ m[/tex].

Explanation:

Given that,

Shearing force F = 600 N

Shear modulus [tex]S = 1\times10^{9}\ N/m^2[/tex]

length = 0.700 cm

diameter = 4.00 cm

We need to find the shear deformation

Using formula of shear modulus

[tex]S=\dfrac{Fl_{0}}{A\Delta x}[/tex]

[tex]\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}[/tex]

[tex]\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}[/tex]

Put the value into the formula

[tex]\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}[/tex]

[tex]\Delta x=3.34\times10^{-6}\ m[/tex]

Hence, The shear deformation is [tex]\Delta x=3.34\times10^{-6}\ m[/tex].

The shear deformation experienced by the disc is calculated using a formula that takes into account the shear modulus, the force applied, and the cross-sectional area of the disk. The correct answer is found to be approximately 0.478 μm.

Solving this problem involves understanding the formula for shear deformation, which is the ratio of the applied force to the area of the disc over which it is applied, multiplied by the height of the disc and divided by the shear modulus.

First, we need to calculate the cross-sectional area of the disk. The formula for the area of a circle is πr², where r is the radius of the disc. Given the diameter of 4 cm, the radius is 2 cm or 0.02 m. So, the area = π * (0.02)² = 0.001256 m².

Substituting into the formula for shear deformation, we get τ = F / (G * A) which equals 600 N / (1x10^9 N/m² * 0.001256 m²) = 4.78x10^-7 m or approximately 0.478 μm.

This indicates that none of the initial answers are correct. The closest incorrect answer is 3 μm but the correct answer is 0.478 μm.

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Answer:

The wave length is [tex]3.885\times10^{-13}\ m[/tex]

Explanation:

Given that,

Energy = 10 Mev

We need to calculate the wavelength

Using formula of debroglie wave length

[tex]\lambda=\dfrac{h}{\sqrt{2mE}}[/tex]

Where, h = Planck constant

E = energy

m = mass

Put the value into the formula

[tex]\lambda =\dfrac{6.634\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times10\times10^{6}\times1.6\times10^{-19}}}[/tex]

[tex]\lambda=3.885\times10^{-13}\ m[/tex]

Hence, The wave length is [tex]3.885\times10^{-13}\ m[/tex]

Answer:

9.1 x 10⁵ ohm

Explanation:

C = Capacitance of the capacitor = 9 x 10⁻⁶ F  

V₀ = Voltage of the battery = 13 Volts  

V = Potential difference across the battery after time "t" = 4 Volts  

t = time interval = 3 sec  

T = Time constant

R = resistance  

Potential difference across the battery after time "t" is given as  

[tex]V = V_{o} (1-e^{\frac{-t}{T}})[/tex]

[tex]4 = 13 (1-e^{\frac{-3}{T}})[/tex]

T = 8.2 sec  

Time constant is given as  

T = RC  

8.2 = (9 x 10⁻⁶) R  

R = 9.1 x 10⁵ ohm

Final answer:

To determine the resistance R, the RC circuit charging equation is used with the given values. By rearranging the equation and solving, the resistance R is found to be approximately 7.97 kΩ.

Explanation:

To find the resistance R in the given circuit, we use the charging equation for a capacitor in an RC circuit:

V(t) = V_0(1 - e^{-t/RC})

Where V(t) is the voltage across the capacitor at time t, V_0 is the initial voltage provided by the battery, R is the resistance, C is the capacitance, and t is the time.

Plugging in the given values:

V(t) = 4.00 V

V_0 = 13.0 V

C = 9.0 µF

t = 3.00 s

We have:

4.00 = 13.0(1 - e^{-3/(9.0×10^{-6}R)})

Now solve for R:

1 - \frac{4.00}{13.0} = e^{-3/(9.0×10^{-6}R))}

Simplifying:

\frac{9.00}{13.00} = e^{-3/(9.0×10^{-6}R))}

Take the natural logarithm of both sides:

ln(\frac{9.00}{13.00}) = -\frac{3}{9×10^{-6}R}

Multiply by -9×10^{-6}R and divide by 3:

R = -\frac{9×10^{-6}ln(\frac{9.00}{13.00})}{3}

R ≈ 7.97 kΩ

Thus, the resistance R is approximately 7.97 kΩ.

Answer:

F = 0.018 N

Explanation:

Magnetic force between two parallel current carrying wires is given by

[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

here we know that

[tex]i_1 = 34 A[/tex]

[tex]i_2 = 69 A[/tex]

d = 15 cm

L = 5.9 m

now from above formula we can say

[tex]F = \frac{(4\pi \times 10^{-7})(34 A)(69 A)5.9}{2\pi (0.15)}[/tex]

now the force between two wires is given as

[tex]F = 0.018 N[/tex]

Answer:

So police car will overtake the speeder after 5.64 s

Explanation:

Initially the distance between police car and the speeder when police car is about to accelerate

[tex]d = (v_1 - v_2)t[/tex]

[tex]v_1 = 110 km/h = 110 \times \frac{1000}{3600}m/s = 30.55 m/s[/tex]

[tex]v_2 = 95 km/h = 95 \times \frac{1000}{3600}m/s = 26.39 m/s[/tex]

[tex]d = (30.55-26.39)(2) = 8.32 m[/tex]

now we have

now velocity of police with respect to speeder is given as

[tex]v_r = v_2 - v_1 = 26.39 - 30.55 = -4.17 m/s[/tex]

relative acceleration of police car with respect to speeder

[tex]a_r = a = 2 m/s^2[/tex]

now the time taken to cover the distance between police car and speeder is given as

[tex]d = v_i t + \frac{1}{2}at^2[/tex]

[tex]8.32 = -4.17 t + \frac{1}{2}(2)(t^2)[/tex]

[tex]t^2 -4.17 t - 8.32 = 0[/tex]

[tex]t = 5.64 s[/tex]

Answer: t = 7.61s

Explanation: The initial speed of the police's car is Vp = 95 km/h.

The initial speed of the car is Vc = 110km/h

The acceleration of the police's car is 2m/s^2

Now, we should write this the quantities in the same units, so lets write the velocities in meters per second.

1 kilometers has 1000 meters, and one hour has 3600 seconds, so we have that:

Vp = 95*1000/3600 m/s = 26.39m/s

Vc = 110*1000/3600 m/s = 30.56m/s

now, after the police car starts to accelerate, the velocity equation will be now.

The positions of the cars are:

P(t) = (a/2)*t^2 + v0*t + p0

Where a is the acceleration, v0 is the initial velocity, and p0 is the initial position.

We know that the difference in the velocity of the cars is two seconds, so after those the speeding car is:

(30.56m/s - 26.39m/s)*2s = 8.34m/s

Now we can write the position equations as:

Pp = 1m/s*t^2 + 26.39m/s*t + 0

Here i assume that the initial position of the car is at the 0 units in one axis.

Pc = 30.56m/s*t + 8.34m/s

Now we want to find the time at wich both positions are the same, and after that time the police car will go ahead of the speeding car.

1m/s*t^2 + 26.39m/s*t  = 30.56m/s*t + 8.34m/s

t^2 + (26.39 - 30.56)*t - 8.34 = 0

t^2 - 4.15*t - 8.34 = 0

Now we need to solve this quadratic equation:

t = (4.15 +/- √(4.15^2 - 4*1*(-8.34))/2 = (4.15 +/- 7.11)/2

From here we have two solutions, one positive and one negative, and we need to take the positive one:

t = (4.15 + 7.11)/2s = 11.26/2 s= 5.61s

And remember that the police car started accelerating two secnods after that the speeding car passed it, so the actual time is:

t = 5.61s + 2s = 7.61s

Answer: [tex]0.000001475s=1.475\mu s[/tex]

Explanation:

The index of refraction [tex]n[/tex] is a number that describes how fast light propagates through a medium or material.  

Being its equation as follows:  

[tex]n=\frac{c}{v}[/tex] (1)

Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum and [tex]v[/tex] its speed in the other medium.

So, from (1) we can find the velocity at which the light travels and then the time it requires to travel : [tex]v=\frac{c}{n}[/tex] (2)

For medium 1:

[tex]n_{1}=1.33[/tex]

[tex]v_{1}=\frac{c}{n_{1}}[/tex] (3)

[tex]v_{1}=\frac{3(10)^{8}m/s}{1.33}=225563909.8m/s[/tex] (4)

For medium 2:

[tex]n_{2}=1.51[/tex]

[tex]v_{2}=\frac{c}{n_{2}}[/tex] (5)

[tex]v_{2}=\frac{3(10)^{8}m/s}{1.51}=198675496.7m/s[/tex] (6)

On the other hand, the velocity [tex]v[/tex] is the distance [tex]d[/tex] traveled in a time [tex]t[/tex]:

[tex]v=\frac{d}{t}[/tex] (7)

We can isolate [tex]t[/tex] from (7) and find the value of the required time:

[tex]t=\frac{d}{v}[/tex] (8)

In this case the total time will be:

[tex]t=t_{1}+t_{2}=\frac{d_{1}}{v_{1}}+\frac{d_{2}}{v_{2}}[/tex] (9)

Where:

[tex]d_{1}=331cm=3.31m[/tex] is the distance the light travels in medium 1

[tex]d_{2}=151cm=1.51m[/tex] is the distance the light travels in medium 2

[tex]v_{1}=225563909.8m/s[/tex] is the velocity of light in medium 1

[tex]v_{2}=198675496.7m/s[/tex] is the velocity of light in medium 2

[tex]t=t_{1}+t_{2}=\frac{3.31m}{225563909.8m/s}+\frac{1.51m}{198675496.7m/s}[/tex] (10)

Finally:

[tex]t=0.000001475s=1.475(10)^{-6}s=1.475\mu s[/tex] (10)

Light takes different amounts of time to travel through different media due to refraction. The time can be calculated by dividing the distance traveled in each medium by the speed of light in that medium.

When light travels from one medium to another, it changes direction, a phenomenon called refraction. The time it takes for light to travel from point A to point B in this case can be calculated by dividing the distance traveled in each medium by the speed of light in that medium. In medium 1, the distance traveled is 331 cm and the index of refraction is 1.33. In medium 2, the distance traveled is 151 cm and the index of refraction is 1.51.

Using the equation time = distance / speed, we can calculate the time it takes for light to travel in each medium.

In medium 1: time1 = 331 cm / speed1

In medium 2: time2 = 151 cm / speed2

Answer:

[tex]d_{o}[/tex] = 154 cm

[tex]d_{i}[/tex] = 216.6 cm

The image is real

Explanation:

[tex]h_{o}[/tex] = height of the object = 3.20 cm

[tex]h_{i}[/tex] = height of the image = 4.50 cm

f = focal length of the converging lens = 90 cm

[tex]d_{o}[/tex] = object distance from the lens = ?

[tex]d_{i}[/tex] = image distance from the lens = ?

using the equation for magnification

[tex]\frac{h_{i}}{h_{o}}= \frac{ d_{i}}{d_{o}}[/tex]

[tex]\frac{4.50}{3.20}= \frac{d_{i}}{d_{o}}[/tex]

[tex]d_{i}[/tex] = 1.40625 [tex]d_{o}[/tex]                        eq-1

using the lens equation

[tex]\frac{1}{d_{i}} + \frac{1}{d_{o}} = \frac{1}{f}[/tex]

using eq-1

[tex]\frac{1}{( 1.40625)d_{o}} + \frac{1}{d_{o}} = \frac{1}{90}[/tex]

[tex]d_{o}[/tex] = 154 cm

Using eq-1

[tex]d_{i}[/tex] = 1.40625 [tex]d_{o}[/tex]  

[tex]d_{i}[/tex] = 1.40625 (154)

[tex]d_{i}[/tex] = 216.6 cm

The image is real

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0²=5.6² + 2 x -9.81 x s

          s = 1.60 m

  Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0= 5.6 - 9.81 x t

          t = 0.57s

  Now considering horizontal motion of skateboarder.  

  We have equation of motion s =ut + 0.5 at²

          u = 6.6cos58 = 3.50 m/s

          a = 0 m/s²

          t = 0.57  

  Substituting

         s =3.5 x 0.57 + 0.5 x 0 x 0.57²

         s = 2 m

  When he reaches maximum height he is 2 m far from end of the ramp.

The highest point reached by the skateboarder is 1.612 meters above the ground. When the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is  3.568 meters.

Given:

Initial velocity (v₀) = 6.6 m/s

Launch angle (θ) = 58°

Height of the ramp (h) = 1.2 m

Acceleration due to gravity (g) = 9.8 m/s²

(a) To find the maximum height reached by the skateboarder:

Δy = v₀y² / (2g)

v₀y = v₀ × sin(θ)

v₀y = 6.6 × sin(58°)

v₀y = 5.643 m/s

Δy = (5.643 )² / (2 × 9.8)

Δy ≈ 1.612 m

Therefore, the highest point reached by the skateboarder is 1.612 meters above the ground.

(b) The time of flight can be calculated using the equation:

t = 2 × v₀y / g

t = 2 × 5.643  / 9.8

t = 1.153 s

Δx = v₀x × t

First, we need to find the initial horizontal velocity (v₀x):

v₀x = v₀ × cos(θ)

v₀x = 6.6 × cos(58°)

v₀x = 3.099 m/s

Δx = 3.099 * 1.153

Δx = 3.568 m

Therefore, when the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is  3.568 meters.

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Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

[tex]a=\dfrac{P}{m}[/tex]

[tex]a=\dfrac{181\ N}{68\ kg}[/tex]

[tex]a=2.67\ m/s^2[/tex]

(c) P = 352 N

[tex]a=\dfrac{P}{m}[/tex]

[tex]a=\dfrac{352\ N}{68\ kg}[/tex]

[tex]a=5.17\ m/s^2[/tex]

Hence, this is the required solution.

Answer:

Induced emf through a loop of wire is 3.5 V.

Explanation:

It is given that,

Initial magnetic flux, [tex]\phi_i=1.7\ Wb[/tex]

Final magnetic flux, [tex]\phi_f=0.3\ Wb[/tex]

The magnetic flux through a loop of wire decreases in a time of 0.4 s, t = 0.4 s

We need to find the average value of the induced emf. It is equivalent to the rate of change of magnetic flux i.e.

[tex]\epsilon=-\dfrac{\phi_f-\phi_i}{t}[/tex]

[tex]\epsilon=-\dfrac{0.3\ Wb-1.7\ Wb}{0.4\ s}[/tex]

[tex]\epsilon=3.5\ V[/tex]

So, the value of the induced emf through a loop of wire is 3.5 V.

Answer:

[tex]V_{rex}=75.65m/s[/tex] and [tex]a_{res}=0[/tex] at t=5 secs

Explanation:

We have r =2m

[tex]\therefore \frac{dr}{dt}=0\\\\=>V_{r}=0[/tex]

Similarly

[tex]=>V_{\theta }=\omega r\\\\\omega =\frac{d\theta }{dt}=\frac{d(\pi t)}{dt}=\pi \\\\\therefore V_{\theta }=\pi r=2\pi[/tex]

Similarly

[tex]=>V_{z }=\frac{dz}{dt}\\\\V_{z}=\frac{dsin(24\pi t)}{dt}\\\\V_{z}=24\pi cos(24\pi t)[/tex]

Hence

at t =5s [tex]V_{\theta}=2\pi m/s[/tex]

[tex]V_{z}=24\pi cos(120\pi)[/tex]

[tex]V_{z}=24\pi m/s[/tex]

[tex]V_{res}=\sqrt{V_{\theta }^{2}+V_{z}^{2}}[/tex]

Applying values we get

[tex]V_{res}=75.65m/s[/tex]

Similarly

[tex]a_{\theta }=\frac{dV_{\theta }}{dt}=\frac{d(2\pi) }{dt}=0\\\\a_{z}=\frac{d^{2}(sin(24\pi t))}{dt^{2}}\\\\a_{z}=-24^{2}\pi^{2}sin(24\pi t)\\\\\therefore t=5\\a_{z}=0[/tex]

Explanation:

(1) The force that exists between charged particles is electrostatic force. It is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Where

q₁ and q₂ are charges

r is distance between charges

If the charge of a particle doubles, the electric force doubles. So, the correct option is (a) "It doubles".

(2) A charge exerts a negative force on another charge. Negative force denotes the force is attractive. It means that the charges are of opposite sign. So, the correct option is (c) "the charges are of opposite signs".

Explanation:

Given that,

Height of object = 4.31 cm

Distance of the object = -12.6 cm

Distance of the image = -8.77 cm

For concave mirror,

Using mirror's formula

[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f}=\dfrac{1}{-12.6}-\dfrac{1}{8.77}[/tex]

[tex]\dfrac{1}{f}=-\dfrac{10685}{55251}[/tex]

[tex]f=-\dfrac{55251}{10685}[/tex]

[tex]f = -5.17\ cm[/tex]

Radius of the mirror is

[tex]f = |\dfrac{R}{2}|[/tex]

[tex]r=2f[/tex]

[tex]r=2\times5.17[/tex]

[tex]r=10.34\ cm[/tex]

The magnification of the mirror,

[tex]m=-\dfrac{v}{u}[/tex]

[tex]\dfrac{h_{i}}{h_{o}}=\dfrac{v}{u}[/tex]

[tex]h_{i}=-h_{o}\times\dfrac{v}{u}[/tex]

[tex]h_{i}=-4.31\times\dfrac{8.77}{12.6}[/tex]

[tex]h_{i}=-2.99\ cm[/tex]

Now, For convex mirror,

Using mirror's formula

[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f}=\dfrac{1}{-12.6}+\dfrac{1}{8.77}[/tex]

[tex]\dfrac{1}{f}=\dfrac{1915}{55251}[/tex]

[tex]f=\dfrac{55251}{1915}[/tex]

[tex]f = 28.85\ cm[/tex]

Radius of the mirror is

[tex]f = \dfrac{R}{2}[/tex]

[tex]r=2f[/tex]

[tex]r=2\times28.85[/tex]

[tex]r=57.7\ cm[/tex]

The magnification of the mirror,

[tex]m=-\dfrac{v}{u}[/tex]

[tex]\dfrac{h_{i}}{h_{o}}=\dfrac{v}{u}[/tex]

[tex]h_{i}=-h_{o}\times\dfrac{v}{u}[/tex]

[tex]h_{i}=4.31\times\dfrac{8.77}{12.6}[/tex]

[tex]h_{i}=2.99\ cm[/tex]

Hence, This is the required solution.

By using equations from physics pertaining to projectile motion and manipulation of initial velocity, final velocity components, and combined final velocity, we can calculate the initial speed at which the rock was thrown and its velocity just before striking the ground.

The subject of this question is projectile motion, a branch of physics. Given that the horizontal range of the throw is equal to the height of the building, we can apply the equation for range in projectile motion: R = (v² sin 2α) / g, where R is the range (20m), v is velocity, α is the angle (53 degrees), and g is acceleration due to gravity (approx. 9.8 m/s²).

Firstly, solve the equation for v which gives v = sqrt(R * g / sin 2α). This gives the starting speed of the rock.

To find the final velocity just before hitting the ground, we need to find vertical and horizontal components of velocity. Vertical component can be obtained by using: v_f = sqrt(v_i² + 2*g*h), h is the height(20m). Horizontal component remains constant which is v_i*cosα. The final velocity is then, sqrt(v_h² + v_f²).

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The speed of the car on Curve B is obtained by solving for the radius in Curve A's equation, then substituting that into the formula for Curve B. The formula involved is based on principles of circular motion and forces.

The subject of this question is Physics, specifically the principles of circular motion and the forces at play within. Given the information from Curve A, we can ascertain that the speed of the car on Curve B can be found using principles of physics. The formula to find the speed at which the car can travel around a banked curve without relying on friction is: v = sqrt[rgtan(Θ)]. Where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and Θ is the angle of the banked curve.

Since the radius is the same for both curves, and the speed is known for Curve A, we can set it up so: 19.1 = sqrt[r * 9.8 * tan(12.7)]. We then solve for r (the radius), and apply it to Curve B: v = sqrt[r * 9.8 * tan(15.1)]. By substituting in the value of r obtained from the first equation, we can calculate the speed for Curve B.

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To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.

The speed at which a car can travel around a banked curve without relying on friction can be calculated using the ideal banking angle formula. The formula is given by v = √(g * r * tan(θ)), where v is the speed, g is the acceleration due to gravity, r is the radius of the curve, and θ is the banked angle. In this case, curve A is banked at 12.7° and the car can travel at a speed of 19.1 m/s. To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.

Plugging in the values into the formula, v = √(9.8 * r * tan(15.1)). Since the radius is the same for both curves, we can solve for v: 19.1 = √(9.8 * r * tan(12.7))

Solving the equation for v, we find that the car can travel around curve B without relying on friction at a speed of approximately 21.2 m/s.

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Answer:

The diameter of a square link is 0.0233 inch.

Explanation:

Given that,

Load = 27000 lbs

Modulus of elasticity [tex]E= 30\times10^{6}\ psi[/tex]

Working stress [tex]\sigma=7000\ psi[/tex]

length l = 55 in

We need to calculate the diameter of a square link

Using formula of stress

[tex] \sigma=\dfrac{Force}{Area}[/tex]

[tex]7000=\dfrac{27000}{\pi\times d\times L}[/tex]

Put the value into the formula

[tex]d=\dfrac{27000}{7000\times3.14\times55}[/tex]

[tex]d=0.0223\ inch[/tex]

Hence, The diameter of a square link is 0.0233 inch.

(a) The velocity of the car-truck system after collision is 18.08 m/s.

(b) The increase in internal energy of the car and truck is 2,835,031.2 J.

The velocity of the system after collision is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(36) + 4000(-15) = vx(2800 + 4000)

40,800 = 6800vx

vx = 6 m/s

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(0) + 4000(29) = vz(2800 + 4000)

116,000 = 6800vz

vz = 17.06 m/s

K.E(car) = ¹/₂mv²

K.E(car) = ¹/₂ x (2800) x (36)²

K.E(car) = 1,814,400 J

v(truck) = √(15² + 29²) = 32.65

K.E(truck) = ¹/₂ x (4000) x (32.65)²

K.E(truck) = 2,132,045 J

K.E(total) =  1,814,400 J + 2,132,045 J = 3,946,445 J

K.E =  ¹/₂(m₁ + m₂)v²

K.E = ¹/₂ x (2800 + 4000) x (18.08)²

K.E = 1,111,413.8 J

U = ΔK.E

U = 3,946,445 J - 1,111,413.8 J

U = 2,835,031.2 J

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Final answer:

To determine the velocity of the stuck-together car and truck after the collision, apply the principle of conservation of momentum. Consider the change in kinetic energy and the production of thermal energy and deformation.

Explanation:

To determine the velocity of the stuck-together car and truck just after the collision, we can apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. We can find the final velocity by summing the momenta of the car and truck and dividing by their combined mass.

In this case, the car's momentum is the product of its mass and velocity, and the truck's momentum is the product of its mass and velocity. Adding these momenta together and dividing by the combined mass gives us the final velocity of the stuck-together car and truck just after the collision.

For the increase in internal energy of the car and truck, we need to consider the change in kinetic energy and the production of thermal energy and deformation. The change in kinetic energy can be calculated by finding the difference between the initial and final kinetic energies of the car and truck. The thermal energy and deformation depend on factors such as the materials involved and the severity of the collision.

Answer:

Part a)

Q = 3198 J

Part b)

It is compression of gas so this is energy transferred to the gas

Explanation:

Part a)

Energy transfer during compression of gas is same as the work done on the gas

In isothermal process work done is given by the equation

[tex]W = nRT ln(\frac{V_2}{V_1})[/tex]

now we know that

n = 1.4 moles

T = 27 degree C = 300 K

[tex]V_2 = 2.5 m^3[/tex]

[tex]V_1 = 1 m^3[/tex]

now we have

[tex]W = (1.4)(8.31)(300)(ln\frac{2.5}{1})[/tex]

[tex]Q = 3198 J[/tex]

Part b)

It is compression of gas so this is energy transferred to the gas

Answer:

70 Hz

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.  

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

When the car is approaching you:

fr = 76 Hz

vr = 0 m/s

When the car is moving away from you:

fr = 65 Hz

vr = 0 m/s

c, vs, and fs are constant.

We can write two equations:

76 = fs c / (c − vs)

65 = fs c / (c + vs)

If we divide the two equations:

76/65 = [fs c / (c − vs)] / [fs c / (c + vs)]

76/65 = [fs c / (c − vs)] × [(c + vs) / (fs c)]

76/65 = (c + vs) / (c − vs)

76 (c − vs) = 65 (c + vs)

76c − 76vs = 65c + 65vs

11c = 141vs

vs = 11/141 c

Substitute into either equation to find fs.

65 = fs c / (c + 11/141 c)

65 = fs c / (152/141 c)

65 = 141/152 fs

fs = 70 Hz

The question involves the Doppler effect in sound waves. To find the original frequency of the car's horn, the mean of the frequencies heard when the car was approaching and receding is calculated. This gives a result of 70.5 Hz.

This question involves the Doppler effect, which is a change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, the source of the sound is the car's horn.

To calculate the actual frequency of the car's horn, you need to take the frequency you heard when the car was approaching (76 Hz) and when it was leaving (65 Hz) and find the mean of these two values. So, the frequency of the car horn is ((76+65)/2) = 70.5 Hz.

This calculation assumes that your movement is minimal compared to that of the car. As such, most of the perceived frequency change is due to the motion of the car, not the observer. Therefore, the actual frequency of the horn is somewhat between the heard frequencies when the car was approaching and receding. This happens because of the change in relative velocity between the source of sound (car) and the observer when the car goes by.

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Answer:

980 dyne

Explanation:

Volume = 1 cm^3, d = 0.92 g / cm^3, D = 1 g/cm^3

In the equilibrium condition, the buoyant force is equal to the weight of the block.

Buoyant force = Volume of block x density of water x g

Buoyant force = 1 x 1 x 980 = 980 dyne