A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) B cos(ot)ż. A circular loop of wire, of radius a/2 and resistance R, is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time. 6.

Answer:

[tex]B = \frac{\mu_0 \rho r^2 \omega}{2}[/tex]

Explanation:

Let the position of the point where magnetic field is to be determined is at distance "r" from the axis of cylinder

so here total charge lying in this region is

[tex]q = \rho(\pi r^2 L)[/tex]

now magnetic field inside the cylinder is given as

[tex]B = \frac{\mu_0 N i}{L}[/tex]

here current is given as

[tex]i = \frac{q\omega}{2\pi}[/tex]

[tex]i = \frac{\rho (\pi r^2 L) \omega}{2\pi}[/tex]

[tex]i = \frac{\rho r^2 L \omega}{2}[/tex]

now magnetic field is given as

[tex]B = \frac{\mu_0 \rho r^2 L \omega}{2L}[/tex]

[tex]B = \frac{\mu_0 \rho r^2 \omega}{2}[/tex]

Answer:

The function has a maximum in [tex]x=3[/tex]

The maximum is:

[tex]f(3) = 39[/tex]

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

[tex]f(x)' = -4*2x + 24=0[/tex]

[tex]-4*2x + 24=0[/tex]

[tex]8x=24[/tex]

[tex]x=3[/tex]

Now find the second derivative of the function and evaluate at x = 3.

If [tex]f (3) ''< 0[/tex] the function has a maximum

If [tex]f (3) '' >0[/tex] the function has a minimum

[tex]f(x)''= 8[/tex]

Note that:

[tex]f(3)''= -8<0[/tex]

the function has a maximum in [tex]x=3[/tex]

The maximum is:

[tex]f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39[/tex]

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:

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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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:

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:

[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:

[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:

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

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.

Answer:

365°C

Explanation:

°C=1.10*23.2/0.0700

°C=365°C//

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:

i = 323 A

Explanation:

Initial flux due to magnetic field from the coil is given as

[tex]\phi = NB.A[/tex]

here we will have

[tex]N = 190 [/tex]

[tex]B = 1.60 T[/tex]

[tex]A = 0.340 m^2[/tex]

now the flux is given as

[tex]\phi_1 = (190)(1.60)(0.340) = 103.36 T m^2[/tex]

finally current in the electromagnet changed to zero

so final flux in the coil is zero

[tex]\phi_2 = 0[/tex]

now we know that rate of change in flux will induce EMF in the coil

so we will have

[tex]EMF = \frac{\phi_1 - \phi_2}{\Delta t}[/tex]

[tex]EMF = \frac{103.36 - 0}{20 \times 10^{-3}}[/tex]

[tex]EMF = 5168 Volts[/tex]

now induced current is given as

[tex]i = \frac{EMF}{R}[/tex]

[tex]i = \frac{5168}{16} = 323 A[/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:

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.

Answers:  

a) How high does the rock get?=1.933m

b)How far downrange from the pedestal does the rock land?=21.25m

Explanation:

This situation is a good example of projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being their main equations as follows:  

x-component:  

[tex]x=V_{o}cos\theta t[/tex]   (1)  

[tex]V_{x}=V_{o}cos\theta[/tex]   (2)  

Where:  

[tex]V_{o}=18m/s[/tex] is the rock's initial speed  

[tex]\theta=20\°[/tex] is the angle

[tex]t[/tex] is the time since the rock is propelled until it hits the ground  

y-component:  

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (3)  

[tex]V_{y}=V_{o}sin\theta-gt[/tex]   (4)  

Where:  

[tex]y_{o}=10m[/tex]  is the initial height of the rock

[tex]y=0[/tex]  is the final height of the rock (when it finally hits the ground)  

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Knowing this, let's begin with the anwers:

Here we are talking about the maximun height [tex]y_{max}[/tex] the rock has in its parabolic motion. This is fulfilled when [tex]V_{y}=0[/tex].

Rewritting (4) with this condition:

[tex]0=V_{o}sin\theta-gt[/tex]   (5)  

Isolating [tex]t[/tex]:

[tex]t=\frac{V_{o}sin\theta}{g}[/tex]  (6)  

Substituting (6) in (3):

[tex]y_{max}=y_{o}+V_{o}sin\theta(\frac{V_{o}sin\theta}{g})-\frac{1}{2}g(\frac{V_{o}sin\theta}{g})^{2}[/tex]   (7)  

[tex]y_{max}=\frac{V_{o}^{2}sin^{2}\theta}{2g}[/tex]   (8)  

Solving:

[tex]y_{max}=\frac{(18m/s)^{2}sin^{2}(20\°)}{2(9.8m/s^{2})}[/tex]   (9)  

Then:

[tex]y_{max}=1.933m[/tex]   (10) This is the maximum height the rock has.

Here we are talking about the maximun horizontal distance [tex]x_{max}[/tex] the rock has in its parabolic motion (this is fulfilled when [tex]y=0[/tex]):

[tex]0=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (11)  

Isolating [tex]t[/tex] from (11):

[tex]t=\frac{2V_{o}sin\theta}{g}[/tex] (12)  

Substituting (12) in (1):

[tex]x_{max}=V_{o}cos\theta (\frac{2V_{o}sin\theta}{g})[/tex]   (13)

[tex]x_{max}=\frac{V_{o}^{2}(2cos\theta sin\theta)}{g}[/tex]   (14)

Knowing [tex]sin(2\theta)=2cos\theta sin\theta[/tex]:

[tex]x_{max}=\frac{V_{o}^{2}sin2\theta}{g}[/tex]   (15)

Solving:

[tex]x_{max}=\frac{(18m/s)^{2}sin2(20)}{9.8m/s^{2}}[/tex]   (16)

Finally:

[tex]x_{max}=21.25m[/tex]   (17)

Answer:

[tex]E = 2 \times 10^{-3} V[/tex]

Explanation:

As we know that rate of change in flux will induce EMF

So here we can

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we have

[tex]EMF = \pi r^2\frac{dB}{dt}[/tex]

now we also know that induced EMF is given by

[tex]\int E. dL = \pi r^2\frac{dB}[dt}[/tex]

[tex]E (2\pi r) = \pi r^2\frac{dB}{dt}[/tex]

[tex]E = \frac{r}{2}(\frac{dB}{dt})[/tex]

now plug in all values in it

[tex]E = \frac{0.0133}{2}(0.12 t)[/tex]

[tex]E = 8 \times 10^{-4} (2.50) = 2 \times 10^{-3} V/m[/tex]

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:

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:

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:

232 J

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = m C (T − T₀)

Given m = 75.1 g, C = 0.14 J/g/°C, T = 53.8°C, and T₀ = 31.7°C:

q = (75.1 g) (0.14 J/g/°C) (53.8°C − 31.7°C)

q = 232 J

Answer:

[tex]Q=232.36J[/tex]

Explanation:

The heat capacity (C) of a physical system depends on the amount of substance of that system. For a system formed by a single homogeneous substance, it is defined as:

[tex]C=mc(1)[/tex]

Here m is the mass of the system and c is the specific heat capacity.

The heat capacity is defined as the ratio between the heat absorbed by the system and the resulting temperature change:

[tex]C=\frac{Q}{\Delta T}(2)[/tex]

We equal (1) and (2) and solve for Q:

[tex]\frac{Q}{\Delta T}=mc\\Q=mc\Delta T\\Q=mc(T_f-T_i)\\Q=75.1g(0.14\frac{J}{g^\circ C})(53.8^\circ C-31.7^\circ C)\\Q=232.36J[/tex]

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.

Answer:

3.7 x 10⁷ m/s

Explanation:

ΔV = Potential difference through which the proton moves = 7.15 MV = 7.15 x 10⁶ Volts

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

v = speed of the proton as it reach zero potential region

m = mass of the proton = 1.66 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy lost

(0.5) m v² = q ΔV

(0.5) (1.66 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.15 x 10⁶)

v = 3.7 x 10⁷ m/s

Final answer:

The speed of protons in a Van de Graaff accelerator transitioning from a 7.15 MV potential to zero potential is about 3.70 x 10^7 m/s, option c

Explanation:

Protons being accelerated in a Van de Graaff accelerator from a 7.15 MV potential to zero potential can be analyzed using the principle of conservation of energy.

The kinetic energy gained by the protons equals the initial potential energy they had, leading to the formula: 1/2 mv^2 = qV.

Calculating this, the speed of the protons when they reach the zero potential region is approximately 3.70 x 10^7 m/s (C).

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:

7.07 x 10⁸ N/m²

Explanation:

F = Force applied to the wire = 68 N

d = diameter of the wire = 0.7 mm = 0.7 x 10⁻³ m

r = radius of the wire = (0.5) d = (0.5) (0.7 x 10⁻³) = 0.35 x 10⁻³ m

Area of cross-section of wire is given as

A = (0.25) πr²

A = (0.25) (3.14) (0.35 x 10⁻³)²

A = 9.61625 x 10⁻⁸ m²

Tensile stress is given as

[tex]P = \frac{F}{A}[/tex]

[tex]P = \frac{68}{9.61625\times 10^{-8}}[/tex]

P = 7.07 x 10⁸ N/m²

Given:

Applied Force on wire = 68 N

Diameter of wire, d = 0.7 mm = [tex]0.7\times 10^{-3}[/tex] m

Radius of wire, r = [tex]\frac{d}{2}[/tex] = 0.35 mm = [tex]0.35\times 10^{-3}[/tex] m

Formula used:

Stress =  [tex]\frac{Applied Force}{cross-sectional area}[/tex]

Explanation:

Cross-sectional area, A = [tex]\pi r^{2}[/tex]  =  [tex]\pi (0.35\times 10^{-3})^{2}[/tex]

A = [tex]3.84\times 10^{-7} m^{2}[/tex]

Using the formula for stress:

Stress =  [tex]\frac{68}{3.84\times 10^{-7}}[/tex] =  [tex]1.76\times 10^{8}[/tex]

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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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:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

The mass of water that needs to evaporate from the skin to reduce a body temperature by 0.45°C is calculated using the body's specific heat capacity, body mass and the latent heat of vaporization of water. First, we find the energy required to cool the body by the temperature change and then find the mass of water that would embody that amount of energy.

The question is asking for the calculation of the mass of water that should evaporate from the skin to reduce the body temperature by 0.45°C. To find this, we need to first understand that evaporation is a main method for body cooling, and it involves a considerable amount of energy being taken from the skin as water changes into vapor.

The energy for evaporating water is explained by the equation Qv = m * Lv where Qv stands for heat energy, m represents mass and Lv is the latent heat of vaporization of water. Given that the specific heat capacity of the body is 4.0 J/g °C and the body mass is 95 kg, the amount of energy required to cool the body by 0.45°C is calculated by multiplying these values (body mass in grams * temperature change in °C * specific heat capacity).

After calculating this energy, we get how much heat needs to be removed from the body to achieve the desired temperature reduction. Lastly, to find the mass of water to be evaporated, we use the equation Qv = m * Lv again but rearrange it as m = Qv / Lv (as Lv = 2256 kJ/kg). This gives us the amount of water that needs to evaporate from the body to reduce the temperature by 0.45°C.

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