If the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.5 m standing 9.0 m away?

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

Answer:

By Newton's law Friction= -62.5N

Explanation:

Applying second law of Newton (which is the the sum of the forces is equal to the mass of the object who feels the forces times the acceleration)

Push+Friction= [tex]m_{box}a_{box}[/tex]

[tex]100+Friction=25*1.5=37.5\\Friction=37.5-100=-62.5 N[/tex]

Which indicates that the friction is opposed to the push and it is really appreciable, that is to say, there is a lot of friction between the box and the floor.

Final answer:

The best approximation of the magnitude of the friction force on the box, when a 25 kg box is accelerated by a 100 N force at 1.5 m/s^2, is 62.5 N.

Explanation:

The student's question involves understanding Newton's second law of motion and the concept of friction. Since a horizontal force of 100 N is applied to a 25 kg box, causing it to accelerate at a rate of 1.5 m/s2, we can calculate the net force using Newton's second law (F = ma). The net force is the difference between the applied force (100 N) and the friction force which we are trying to find.

First, calculate the net force:
Fnet = m × a = 25 kg × 1.5 m/s2 = 37.5 N

Next, calculate the friction force:
Ffriction = Fapplied - Fnet = 100 N - 37.5 N = 62.5 N

Therefore, the best approximation of the magnitude of the friction force on the box is 62.5 N.

Answer:

(a):  345.6 N.

(b): [tex]\rm 2.069\times 10^{29}\ m/s^2.[/tex]

Explanation:

Given:

According to Coulomb's law, the electric field due to a charged sphere of charge Q at a point r distance away from its center is given by

[tex]\rm E=\dfrac{kQ}{r^2}.[/tex]

where, k is the Coulomb's constant, having value = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex]

Therefore, the electric field due to the 125 Xe nucleus at the proton is given by

[tex]\rm E=\dfrac{kq}{r^2}=\dfrac{(9\times 10^9)\times (+54 e)}{r^2}[/tex]

Here,

e is the elementary charge, having value = [tex]\rm 1.6\times 10^{-19}\ C.[/tex]

r is the distance of the proton from the center of the nucleus = [tex]\rm a + \dfrac d2 = 3.0+\dfrac{6.0}2=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]

Using these values,

[tex]\rm E=\dfrac{(9\times 10^9)\times (+54 \times 1.6\times 10^{-19})}{(6.0\times 10^{-15})^2}=2.16\times 10^{21}\ N/C.[/tex]

Now, the electric force on a charge q due to an electric field is given as

[tex]\rm F=qE[/tex]

For the proton, [tex]\rm q = e =1.6\times 10^{-19}\ C.[/tex]

Thus, the electric force on the proton is given by

[tex]\rm F = 1.6\times 10^{-19}\times 2.16\times 10^{21}=345.6\ N.[/tex]

According to Newton's second law,

[tex]\rm F=ma[/tex]

where, a is the acceleration.

The mass of the proton is [tex]\rm m_p=1.67\times 10^{-27}\ kg.[/tex]

Therefore, the proton's acceleration is given by

[tex]\rm a = \dfrac{F}{m_p}=\dfrac{345.6}{1.67\times 10^{-27}}=2.069\times 10^{29}\ m/s^2.[/tex]

(a) The electric force of the proton is 2.56 x 10⁻²⁹ N.

(b) The acceleration of the proton is 0.0153 m/s².

The electric force of the proton is calculated using Coulomb's law of electrostatic force.

F = kq²/r²

F = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹)/(3)²

F = 2.56 x 10⁻²⁹ N

The acceleration of the proton is determined by applying Newton's second law of motion;

F = ma

a = F/m

a = (2.56 x 10⁻²⁹ )/(1.67 x 10⁻²⁷)

a = 0.0153 m/s²

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Answer:0.21

Explanation:

Given

Two point charges are brought closer together, increasing the force by a factor of 22

Let the original force be

[tex]F=\frac{kq_1q_2}{r^2}---1[/tex]

where [tex]q_1,q_2[/tex] are charges and r is the distance between them

new force [tex]F'=\frac{kq_1q_2}{r'^2}----2[/tex]

divide 1 & 2

[tex]\frac{F'}{F}=\frac{\frac{kq_1q_2}{r'^2}}{\frac{kq_1q_2}{r^2}}[/tex]

[tex]22=\frac{r^2}{r'^2}[/tex]

[tex]r'=\frac{r}{\sqrt{22}}\approx 0.213 r[/tex]

Distance between them is decrease by a factor of 0.21

The separation between two point charges decreased by a factor of 5.

To find the factor by which the separation between two point charges decreased when the force between them increased by a factor of 25, we need to understand the relationship between force and distance. According to Coulomb's law, the force between two point charges is inversely proportional to the square of the distance between them. This means that when the distance decreases, the force increases, and vice versa.

Since the force increased by a factor of 25, we can find the factor by which the distance decreased by taking the square root of 25, which is 5.

Therefore, the separation between the two point charges decreased by a factor of 5.

Answer:

It takes the wolf 3 s to move as fast as the rabbit.

The wolf is 9 m from the rabbit when it reaches a speed of 6 m/s

Explanation:

The equations for the position and velocity of objects moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the object at time t

x0 = initial position of the object

t = time

v0 = initial speed

a = acceleration

v = velocity of the object at time t

For the first question, let´s use the equation of velocity of the wolf to find at which time its velocity is the same as the velocity of the rabbit ( 6 m/s):

v = v0 + a · t       (v0 = 0 because the wolf starts at rest)

6 m/s = 0 + 2.0 m/s² · t

t = 3 s

Now, with this calculated time, let´s obtain the position of the wolf:

x = x0 + v0 · t + 1/2 · a · t²      (Placing the center of the frame of reference at the point when the wolf starts running makes x0 = 0)

x = 1/2 · a · t²

x = 1/2 · 2.0 m/s² · (3 s)²

x = 9 m

Now, let´s calculate the position of the rabbit. Notice that a = 0. Then:

x = x0 + v · t      x0 = 0

x = v · t

x = 6 m/s · 3 s = 18 m

The wolf is (18 m - 9 m) 9 m from the rabbit when it reaches a speed of 6 m/s

Answer:

(a) angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side

b] largest order = 4

Explanation:

(a) for diffraction maxima,

[tex]sin \theta =m\times \lambda/d[/tex]

Here, m is the order, [tex]\lambda[/tex] is the wavelength, [tex]\theta[/tex]  is the angle at which maxima occur, d is inter planar spacing.

And we know that lines per mm (N) is related with d as,

[tex]N=\frac{1}{d}[/tex]

Given that the wavelength is,

[tex]\lambda=600.0 nm=600\times 10^{-9}m[/tex]

And [tex]N=\frac{400 lines}{mm} \\N=\frac{400 lines}{10^{-3}m }[/tex]

Now,

[tex]sin \theta =m\times \lambda\times N[/tex]

Therefore,

[tex]sin \theta= m\times600\times 10^{-9} \times 400\times 10^{3}\\sin \theta=0.24m[/tex]

Here, m can be 1,2,3,4 as sin theta has to be less than 1.

[tex]\theta = arcsin 0.24 , arcsin 0.48 , arcsin 0.72 , arcsin 0.96[/tex]

Therefore, angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side

b] largest order = 4

Final answer:

The angles at which the maxima are observed can be determined using dsinθ = mλ. For a grating with 400 lines per mm and light of wavelength 600.0 nm, the largest order that can be seen occurs when the angle of the next order is equal to or greater than the angle of the first order maximum.

Explanation:

The angles at which the maxima are observed can be determined using the formula:

dsinθ = mλ

Where d is the grating spacing, θ is the angle, m is the order of the maximum, and λ is the wavelength of the light.

For a grating with 400 lines per mm and light of wavelength 600.0 nm:

d = 1/400 mm = 0.0025 mm = 0.0025 cm

θ = sin-1((mλ)/d)

To determine the largest order that can be seen, we need to find the angle at which the next order would overlap with the first order. This would be when the angle of the next order is equal to or greater than the angle of the first order maximum.

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

=[tex]\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}[/tex]

a = 30 units    and  b = 70 units

= [tex]\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}[/tex]

=   40 units to 100 units

hence, the correct answer is option C which is 50 units.

Answer:

8.75 m/s^2

Explanation:

initial speed, u = 300 m/s

final speed, v = 400 m/s

distance, h = 4 km = 4000 m

Let a be the acceleration of the plane.

Use third equation of motion

[tex]v^{2}=u^{2}- 2 gh[/tex]

[tex]400^{2}=300^{2}- 2 \times a \times 4000[/tex]

a = 8.75 m/s^2

Thus, the acceleration of the plane is 8.75 m/s^2.

To return to the tree after finding the treasure, you need to calculate the total displacement vector from the tree to the treasure by adding up the vectors of each leg of the journey. The direction and magnitude of the opposite of this vector will provide the direction and distance to get back to the tree. A graphical representation of the vectors will help confirm this.

The problem is about determining the direction and distance to return to the original location after walking certain distances at specific angles. This involves the understanding of vectors and how they are applied in real life.

Given the three legs of the journey:

Summing up these components for all the legs, we get the total displacement vector from the tree to the treasure. To get back to the tree, you would need to walk in the opposite direction of this vector. The magnitude of this vector gives the total distance to be covered.

Comparing this calculated result with a graphical solution will help validate the projected direction and distance.

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Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

The horizontal distance at which the the package must be dropped and the vertical velocity are; 316.23 m and -22.08 m/s

What is the vertical velocity?

A)To fall 200 m, the time required is given by the formula;

t = √(2H/g)

t= √(200/9.8)

t = 4.518 seconds.

The supplies will travel forward by a distance of;

Δx  = 4.518 * 70

Δx  = 316.23 m

B) Time in the horizontal direction is;

t = Δx/V

t = 316.23/70

t = 4.52 s

In the vertical direction;

y = ut + ¹/₂at²

-200 = 4.52u + ¹/₂(-9.81)(4.52)²

100.21 - 200 = 4.52u

u = -99.79/4.52

u = -22.08 m/s

C) Speed at which the package lands in the latter case is;

v = √(70² + (-22.08²))

v = 73.4 m/s

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Answer: a)

Explanation: The phenomenon of emission is related to electronic transitions with the atom so brightly emission lines can represent the most important electronics transitions. They cover the whole spectrum from UV to IR.

Answer:6.806 MN

Explanation:

Given

Charge on clouds is +44 C and -44 C

They are separated by 1.6 km

and electrostatic Force is given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

[tex]F=\frac{9\times 10^9\times 44\times 44}{1600^2}[/tex]

[tex]F=\frac{17,424\times 10^9}{256\times 10^4}[/tex]

F=6.806 MN

Answer:

Sapphire and diamond impurities change with their transparency and role of impurities in dielectrics is discussed below.

Explanation:

Temperature dependence of the dielectrics with different degrees of purity measured in the range of 233-313 K. This was observed that the  impurities presence is  strongly influences the dielectric constant, dipole moment and melting point.

SAPPHIRE

Sapphire is a very important gem stone. It's color is blue, but the naturally occurring sapphires have  purple, yellow,orange and green. This variety in color is due to trace amount of presence of impurities like Iron,titanium, chromium, copper and Magnesium. It is the third hardest materials in the Mohs scale. The transparency of Sapphire is due to the impurities present in it.

DIAMOND

Diamond is the very precious stone. The transparency in the diamond is due to the band structure, that is the band gap is high in the diamonds, so they are transparent in nature. Diamond has many unequaled qualities and is very unique among the other minerals. It has highest refractive index of any natural minerals and is transparent over greatest number of wavelengths due to conduction present in it.

Using the formula for distance covered in free fall, the coin was determined to have been dropped from a height of 19.6 meters.

To determine the height from which a coin was dropped if it reaches the ground in 2 seconds, we can use the formula for the distance covered in free fall (neglecting air resistance), which is given by:

Distance (d) = ½ * g * t2

where g is the acceleration due to gravity (9.8 m/s2 on Earth), and t is the time in seconds. Plugging in the values, we get:

Distance (d) = ½ * 9.8 * (22) = ½ * 9.8 * 4 = 19.6 meters.

Thus, the coin was dropped from a height of 19.6 meters.

The coin was dropped from a height of 19.62 meters, using the equation [tex]\( s = \frac{1}{2}gt^2 \)[/tex].

To solve this problem, we can use the equation of motion for an object in free fall under gravity:

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

Where:

- [tex]\( s \)[/tex] is the displacement (height) of the object

- [tex]\( u \)[/tex] is the initial velocity (which is 0 because the object is dropped)

- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex] on Earth)

- [tex]\( t \)[/tex] is the time taken

Given that the initial velocity [tex]\( u = 0 \)[/tex], we can simplify the equation to:

[tex]\[ s = \frac{1}{2}gt^2 \][/tex]

Substituting the known values:

[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (2 \, \text{s})^2 \][/tex]

[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times 4 \, \text{s}^2 \][/tex]

[tex]\[ s = \frac{1}{2} \times 39.24 \, \text{m} \][/tex]

[tex]\[ s = 19.62 \, \text{m} \][/tex]

So, the height from which the coin was dropped is [tex]\( 19.62 \, \text{m} \)[/tex].

Answer:

The power radiated by the hawk is 0.452 Watt.

Explanation:

Given that,

Frequency = 50 Hz

Distance r=60 m

Level = 70 dB

We need to calculate the intensity

Using formula of intensity

[tex]dB=10 log(\dfrac{I}{I_{0}})[/tex]

Put the value into the formula

[tex]70=10 log(\dfrac{I}{10^{-12}})[/tex]

[tex]I=10^{7}\times10^{-12}[/tex]

[tex]I = 1\times10^{-5}\ W/m^2[/tex]

We need to calculate the power radiated by the hawk

Using formula of power

[tex]P = I\times 4\pi r^2[/tex]

Put the value into the formula

[tex]P=1\times10^{-5}\times4\pi\times(60)^2[/tex]

[tex]P=0.452\ W[/tex]

Hence, The power radiated by the hawk is 0.452 Watt.

Part A: The magnitude of the electric field at the origin is approximately [tex]57.89 \text{ N/C}[/tex].

Part B: The direction of the net electric field at the origin is approximately [tex]249.44°[/tex] relative to the negative x-axis.

To solve this problem, we need to calculate the net electric field at the origin (0, 0) due to the two-point charges q₁ and q₂. Here’s a step-by-step breakdown of the process:

Given Information:

Charge q₁ = -4.00 nC located at (0.600 m, 0.800 m)

Charge q₂ = +6.00 nC located at (0.600 m, 0.0 m)

Part A: Calculate the Magnitude of the Electric Field at the Origin

Calculate the Distance from Each Charge to the Origin:

For q₁:

[tex]r_1 = \sqrt{(0 - 0.600)^2 + (0 - 0.800)^2} \\= \sqrt{0.36 + 0.64} \\= \sqrt{1.00} \\= 1.00 \text{ m}[/tex]

For q₂:

[tex]r_2 = \sqrt{(0 - 0.600)^2 + (0 - 0.0)^2} \\= \sqrt{0.36} \\= 0.600 \text{ m}[/tex]

Calculate the Electric Field due to Each Charge:

The formula for the electric field due to a point charge is:

[tex]E = \frac{k |q|}{r^2}[/tex]

Where [tex]k \approx 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2[/tex]

For q₁:

[tex]E_1 = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 4.00 \times 10^{-9} \text{ C}}{(1.00)^2} \\= 35.96 \text{ N/C}[/tex]

Direction of E₁ is towards q₁ (since it is negative) and therefore directed along the vector from (0.600, 0.800) to the origin.  The components can be calculated as:

[tex]E_{1x} = -E_1 \cdot \frac{0.600}{1.00} = -35.96 \cdot 0.600 \\= -21.576 \text{ N/C}[/tex]

[tex]E_{1y} = -E_1 \cdot \frac{0.800}{1.00} \\= -35.96 \cdot 0.800 \\= -28.768 \text{ N/C}[/tex]

For q₂:

[tex]E_2 = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 6.00 \times 10^{-9} \text{ C}}{(0.600)^2} \\= 24.99 \text{ N/C}[/tex]

Direction of E₂ is away from the charge because it is positive, directed along the negative y-axis:

[tex]E_{2x} = 0[/tex]

[tex]E_{2y} = -E_2 = -24.99 \text{ N/C}[/tex]

Calculate the Net Electric Field at the Origin:

Combine the x and y components of the electric fields:

[tex]E_{net,x} = E_{1x} + E_{2x} \\= -21.576 + 0 \\= -21.576 \text{ N/C}[/tex]

[tex]E_{net,y} = E_{1y} + E_{2y} \\= -28.768 - 24.99 \\= -53.758 \text{ N/C}[/tex]

The total electric field magnitude is:

[tex]E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2}[/tex]

[tex]E_{net} = \sqrt{(-21.576)^2 + (-53.758)^2} \\= \sqrt{465.33 + 2884.46} \\= \sqrt{3349.79} \\= 57.89 \text{ N/C}[/tex]

Final Answer for Part A: The magnitude of the electric field at the origin is approximately [tex]57.89 \text{ N/C}[/tex].

Part B: Calculate the Direction of the Net Electric Field

Calculate the Angle of the Electric Field:

Final Answer for Part B: The direction of the net electric field at the origin is approximately [tex]249.44°[/tex] relative to the negative x-axis.

Complete Question:

A point charge q1 = −4.00nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00nC is at the point x = 0.600 meters, y = 0.

Part A

Calculate the magnitude E of the net electric field at the origin due to these two-point charges. (Express your answer in newtons per coulomb to three significant figures.)

Part B

What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two-point charges. (Express your answer in degrees to three significant figures).

Answer: [tex]0.1851 \frac{kg}{m^{3}}[/tex]

Explanation:

We have this concentration in units of [tex]mg/ml[/tex]:

[tex]\frac{45 mg}{243 ml}[/tex]

And we need to express it in [tex]kg/m^{3}[/tex], knowing:

[tex]1 ml= 1 cm^{3}[/tex]

[tex]1 m^{3}= (100 cm) ^{3}[/tex]

[tex]1g = 1000 mg[/tex]

[tex]1 kg = 1000 g[/tex]

Hence:

[tex]\frac{45 mg}{243 cm^{3}} \frac{1 g}{1000 mg} \frac{1 kg}{1000 g} \frac{(100 cm) ^{3}}{1 m^{3}}=0.1851 \frac{kg}{m^{3}}[/tex]

The cholesterol concentration of the milk in kilograms per cubic meter is 0.1851 kg/m³

The calculation is done as follows;

1 mg= 10⁻⁶ kg

1ml= 10⁻⁶ m³

= 45 ÷243

= 0.1851 kg/m³

Hence the cholesterol concentration of the milk is 0.1851 kg/m³

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Final answer:

The problem involves a ballistic pendulum scenario to determine the initial speed of a bullet, requiring an analysis that combines conservation of momentum and energy. However, the twist in this problem is that the bullet exits the block, requiring indirect methods to deduce the initial conditions.

Explanation:

The question involves finding the initial speed of a bullet by analyzing a ballistic pendulum scenario. This is a physics problem that integrates concepts of momentum and energy conservation. The solution requires two steps: first, finding the velocity of the bullet-block system right after the bullet exits, and second, using energy conservation to link this velocity to the height reached by the block.

Use the conservation of momentum to find the system's velocity right after the bullet exits the block. The exiting speed of the bullet and the mass details are essential for this step. However, an important observation is that the information given does not allow for the direct application of momentum conservation in the traditional sense, since the bullet exits the block, unlike the typical scenario where it remains lodged in.

Next, use the conservation of energy principle to relate the kinetic energy of the block right after the bullet exits to the potential energy at the maximum height. The height given allows calculation of the velocity of the block right after the bullet exits, which indirectly informs the system's dynamics at impact.

This problem has a twist as the bullet does not remain in the block, which complicates the direct application of the conservation of momentum to find the initial speed of the bullet. Instead, it involves a more nuanced approach using the given exit velocity of the bullet and the rise of the block to infer the initial conditions.

The initial speed of the bullet is approximately 23.68 m/s.

To solve this problem, we can use the principle of conservation of momentum and conservation of energy.

1. Conservation of momentum:

The total momentum before the collision is equal to the total momentum after the collision.

[tex]\[m_b v_b = (m_b + m_p)v'\][/tex]

Where:

[tex]\(m_b = 7.0 \, \text{g} = 0.007 \, \text{kg}\)[/tex] (mass of the bullet)

\(v_b\) = initial speed of the bullet (which we need to find)

[tex]\(m_p = 1.5 \, \text{kg}\)[/tex] (mass of the ballistic pendulum)

\(v'\) = final velocity of the bullet and pendulum system

2. Conservation of energy:

The initial kinetic energy of the bullet is equal to the sum of the final kinetic energy of the bullet and the kinetic energy of the pendulum, plus the gravitational potential energy gained by the pendulum.

[tex]\[ \frac{1}{2} m_b v_b^2 = \frac{1}{2} (m_b + m_p) v'^2 + m_p g h\][/tex]

Where:

[tex]\(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)[/tex]

[tex]\(h = 12 \, \text{cm} = 0.12 \, \text{m}\) (maximum height reached by the pendulum)[/tex]

Let's solve these equations step by step:

Step 1: Conservation of momentum:

[tex]\[0.007 \, \text{kg} \times v_b = (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'\][/tex]

[tex]\[0.007 \, \text{kg} \times v_b = 1.507 \, \text{kg} \times v'\][/tex]

[tex]\[v_b = \frac{1.507}{0.007} \times v' \][/tex]

[tex]\[v_b = 215.28 \times v'\][/tex]  (Equation 1)

Step 2: Conservation of energy:

[tex]\[ \frac{1}{2} \times 0.007 \, \text{kg} \times v_b^2 = \frac{1}{2} \times (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'^2 + 1.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.12 \, \text{m}\][/tex]

[tex]\[0.5 \times 0.007 \, \text{kg} \times v_b^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]

Now, substitute [tex]\(v_b = 215.28 \times v'\)[/tex] from Equation 1 into the above equation:

[tex]\[0.5 \times 0.007 \, \text{kg} \times (215.28 \times v')^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]

Simplify and solve for \(v'\):

[tex]\[0.5 \times 0.007 \times (215.28)^2 \times v'^2 = 0.5 \times 1.507 \times v'^2 + 1.47\][/tex]

[tex]\[ 160.49 \times v'^2 = 0.7535 \times v'^2 + 1.47\][/tex]

[tex]\[159.7365 \times v'^2 = 1.47\][/tex]

[tex]\[v'^2 = \frac{1.47}{159.7365}\][/tex]

[tex]\[v' = \sqrt{\frac{1.47}{159.7365}}\][/tex]

[tex]\[v' \approx 0.11 \, \text{m/s}\][/tex]

Finally, we can use this value of \(v'\) to find \(v_b\) using Equation 1:

[tex]\[v_b = 215.28 \times 0.11\][/tex]

[tex]\[v_b \approx 23.68 \, \text{m/s}\][/tex]

So, the initial speed of the bullet is approximately [tex]\(23.68 \, \text{m/s}\).[/tex]

Final answer:

The sparrow exerts a net force on the air in the downward and backward directions as per Newton's Third Law, as it pushes against the air to get lift and move forward.

Explanation:

The sparrow flying around in a circle at a constant speed and height with air resistance is subject to several forces. However, when we consider the force exerted by the sparrow on the air, we need to apply Newton's Third Law of motion which states that for every action, there is an equal and opposite reaction.

Therefore, as the sparrow's wings push the air downward and backward to get lift and forward movement, the air will exert an equal and opposite force on the sparrow. The sparrow's net force on the air would be downward and backward because the sparrow pushes the air in these directions to maintain flight.

Answer:

Explanation:

[tex]x = k\times a^m\times t^n[/tex]

k is constant , it is dimension is zero. Using dimensional unit , we cal write the relation as follows

L = [tex](LT^{-2})^m(T)^n[/tex]

= [tex]L^mT^{-2m+n}[/tex]

Equating power of like items

m=1

-2m+n = 0

n = 2

Answer:

The electric field strength is [tex]4.5\times 10^{4} N/C[/tex]

Solution:

As per the question:

Area of the electrode, [tex]A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}[/tex]

Charge, q = 50 nC = [tex]50\times 10^{- 9} C[/etx]

Distance, x = 2 mm = [tex]2\times 10^{- 3} m[/tex]

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

[tex]\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}[/tex]

Now, the electric field strength of the electrode is:

[tex]\vec{E} = \frac{\sigma}{2\epsilon_{o}}[/tex]

where

[tex]\epsilon_{o} = 8.85\times 10^{- 12} F/m[/tex]

[tex]\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}[/tex]

[tex]\vec{E} = 4.5\times 10^{4} N/C[/tex]

Answer:

15.579 kW

Explanation:

current passing through the wire

[tex]I = \dfrac{P}{V} = \dfrac{690 \times 10^3}{12000} = 57.5 A[/tex]

power loss

[tex]P_{loss} = I^2R = (57.5)^2\times 5 = 16531.25 W[/tex]

the current transmission for 50,000 V is

[tex]I' = \dfrac{P}{V'} = \dfrac{690 \times 10^3}{50000} = 13.8 A[/tex]

power loss

[tex]P_{loss} = I^2R = (13.8)^2\times 5 = 952.2 W[/tex]

wasted power

   =  16531.25 W - 952.2 W

   = 15579.05 W = 15.579 kW

Hence, the power wasted is equal to 15.579 kW

Electric power delivers from power station is the rate of energy transfer per unit time. The less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.

The electric power is the amount of electric energy transferred per unit time. It can be given as,

[tex]P=IV[/tex]

Here, (I) is the current and (V) is the electric potential difference.

Given information-

The power delivers by the power station is 690 kW.

The voltage of the power is 12,000 V.

The resistance of the wire is 5.0 Ω.

As the power delivers by the power station is 690 kW and the voltage of the power is 12,000 V. Thus the current flowing through the wire is,

[tex]I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{12000}\\I=57.5 \rm A[/tex]

The power loss due to the resistance of the wire is 5.0 Ω is,

[tex]P_{loss}=I^2R\\P_{loss}=(57.5)^2\times5\\P_{loss}=16531.25\rm W\\[/tex]

Now the electricity is delivered at 50,000 V rather than 12,000 V.

The power delivers by the power station is 690 kW and the The voltage of the power is 50,000 V. Thus the current flowing through the wire is,

[tex]I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{50000}\\I=13.8 \rm A[/tex]

The power loss due to the resistance of the wire is 5.0 Ω is,

[tex]P_{loss}=I^2R\\P_{loss}=(13.8)^2\times5\\P_{loss}=952.9\rm W\\[/tex]

As when the voltage of the power is 12,000 V the power loss is 16531.25 W and when the voltage of the power is 50,000 V the power loss is 952.9 W.

Thus less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is,

[tex]P_{saved}=16531.25-952.2\\P_{saved}=15579\rm W\\P_{saved}=15.579\rm kW[/tex]

Hence, the amount of less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.

Learn more about the electric power here;

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Answer:

Metals have more mass per unit of volume than wood.

Explanation:

Density is defined as the amount of matter contained in a unit of volume. A material that is denser than another will therefore have more mass per unit of volume.

The density of a body can affect buoyancy, bodies with low enough densities can float in fluids. For example wood can float in water while metals can't.

Answer: d. 5 m/s^2

Explanation:

Acceleration is the change in velocity in a given time.

a = (30-20)/2 = 5

The ball is travelling in horizontal direction with an acceleration of [tex] \bold{5\ m/ s^{2}} [/tex]

Answer: Option D

Explanation:

The rate at which velocity of an object changes is acceleration i.e., if the velocity of an object is changing, the object is accelerating. The unit of measurement of acceleration is Meter per Seconds Square. Acceleration is the derivative of the changing velocity wrt to time,

[tex]a=\frac{\Delta v}{\Delta t}-----{Eqn} 1[/tex]

Given,  

Initial velocity = 20 m/s [tex](v_{i})[/tex]  

Final velocity = 30 m/s [tex](v_{f})[/tex]

Initial time = 0 s [tex](t_{i})[/tex]

Final time = 2 s [tex](t_{f})[/tex]

From the equation 1,

[tex]\begin{aligned} a &=\frac{\Delta v}{\Delta t}=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} \\ a &=\frac{30 \mathrm{m} / \mathrm{s}-20 \mathrm{m} / \mathrm{s}}{2 \mathrm{s}}=5 \mathrm{m} / \mathrm{s}^{2} \end{aligned}[/tex]

Hence, the acceleration of ball moving horizontally is [tex] \bold{5\ m/s^{2}} [/tex]

Answer:

material A

Explanation:

Specific heat of material A = 100 J/kg - K

Specific heat of material B = 200 J/kg - K

The specific heat of a material is defined as the amount of heat required to raise the temperature of substance of mass 1 kg by 1 °C.

So, material A requires 100 J of heat to raise the temperature of mass 1 kg by 1°C.

So, material B requires 200 J of heat to raise the temperature of mass 1 kg by 1°C.

So, the temperature of material A is higher is more than the material B when same amount of heat is added.

Answer:

A) It takes the truck 8 s to catch the motorcycle.

B) The motorcycle has traveled 160 m in that time.

C) The velocity of the truck is 40 m/s at that time.

Explanation:

The equations of the position and velocity of an object moving in a straight line are as follows:

x = x0 +v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

(A) When the the truck catches the motorcycle, both have the same position. Notice that the motorcycle moves at constant speed so that a = 0:

x truck = x motorcycle

x0 +v0 · t + 1/2 · a · t² = x0 + v · t

Placing the origin of the frame of reference at the point where the truck starts, both have an initial position of 0. The initial velocity of the truck is 0. Then:

1/2 · a · t² = v · t

solving for t:

t = 2 v/a

t = 2 · 20 m/s/ 5 m/s²

t = 8 s

It takes the truck 8 s to catch the motorcycle.

(B) Using the equation of the position of the motorcycle, we can calculate the traveled distance in 8 s.

x = v · t

x = 20 m/s · 8 s

x = 160 m

(C) Now, we use the velocity equation at time 8 s.

v = v0 + a · t

v = 0 m/s + 5 m/s² · 8 s

v = 40 m/s

To solve the problem, we equate the displacements of the truck and motorcycle to find when they catch up to each other. The motorcycle travels a total of 160 meters in 8 seconds, while the truck accelerates to a speed of 40 m/s in that time.

The subject at hand involves concepts of uniform motion and constant acceleration, relevant to the study of Physics. We have a motorcycle moving at a constant speed of 20 m/s and a truck starting from rest and accelerating at a rate of 5 m/s².

To answer part (A) we use the equation for the displacement of each vehicle. Our goal is to find the time 't' where their displacements are equal. For the motorcycle, since it's moving at a constant speed, the equation for displacement is given by: x = Ut, where U = initial velocity is 20 m/s and x = displacement For the truck, since it is accelerating, the equation for displacement is 1⁄2at², where a = acceleration is 5 m/s².

Setting these displacements equal to each other gives: Ut = 1⁄2at² To solve for time 't', we obtain: t = 2U/a = 2(20 m/s)/(5 m/s²) = 8 seconds.

For part (B), we calculate how far the motorcycle has traveled using its equation for displacement: x = Ut = 20 m/s * 8 seconds = 160 meters.

Finally, for part (C), we find the truck's final speed using the equation v = U + at. Since the truck starts from rest, U = 0, thus: v = at = 5 m/s² * 8 seconds = 40 m/s.

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Answer:

The charge on the sphere is [tex]1.735\times 10^{- 7} C[/tex]

Solution:

The electric potential on the surface of a sphere of radius 'b' and charge 'Q' is given by:

[tex]V_{sphere} = \frac{Q}{4\pi\epsilon_{o}b}[/tex]

According to the question:

Diameter, b = 0.8 m

Potential of sphere, [tex]V_{sphere} = 39 kV = 39000 V[/tex]

[tex]{4\pi\epsilon_{o}\frac{0.8}{2}\times 39000 = Q[/tex]

Q =  [tex]1.735\times 10^{- 7} C[/tex]

Answer:

(a) 20 kA (B) [tex]diameter=5.0035\times 10^{-2}m[/tex] (c) length = 29.94 m

Explanation:

We have given power P = 100 KW

Voltage V = 5 V

Current density [tex]10A/m^2=10\times 10^{6}A/m^2[/tex]

(A) We have to calculate current

We know that power P = voltage × current

So [tex]100\times 1000=5\times i[/tex]

[tex]i=20kA[/tex]

(b) We know that current density is given by

Current density [tex]=\frac{current}{area}[/tex]

So [tex]10\times 10^{6}A/m^2=\frac{20\times 10^3}{area}[/tex]

[tex]area=2\times 10^{-3}m^2[/tex]

We know that [tex]a=\pi r^2[/tex]

[tex]2\times 10^{-3}=3.14\times r^2[/tex]

[tex]r^2=6.34\times 10^{-4}[/tex]

[tex]r=2.517\times 10^{-2}m[/tex]

[tex]d=2 r=2\times 2.517\times 10^{-2}=5.035\times 10^{-2}m[/tex]

(c) We have to find the length of the cable

We know that resistance [tex]R=\frac{V}{I}=\frac{5}{20\times 10^3}=0.25mohm[/tex]

Resistance is given by [tex]R=\rho \frac{l}{A}[/tex]

So [tex]0.25\times 10^{-3}=1.67\times 10^{-8}\frac{l}{2\times 10^{-3}}[/tex]

l = 29.94 m

Answer:

a)[tex]r_1=1.36R[/tex]

b)[tex]r_2=2.083R[/tex]

Explanation:

Given:

a) when the initial velocity of the projectile is 0.520 times the escape velocity from the earth.

Let r be the radial distance from the earth's surface Let M be the mass of the Earth and R be the radius of the Earth

Now using conservation of Energy at earths surface and at distance r we have

[tex]\dfrac{-GMm}{R}+\dfrac{m(0.52V_e)^2}{2}=\dfrac{-GMm}{r_1}\\\dfrac{-GMm}{R}+\dfrac{m\times 0.52^2\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_1}\\r_2=1.36\ R[/tex]

b) when the Initial kinetic Energy of the projectile is 0.52 times the Kinetic Energy required to escape the Earth

Conservation of Energy we have

[tex]\dfrac{-GMm}{R}+0.52\times KE_{escape}=\dfrac{-GMm}{r_2}\\\dfrac{-GMm}{R}+0.52\times\dfrac{m\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_2}\\r_2=2.083\ R[/tex]

Answer:

See it in the pic

Explanation:

See it in the pic

Final answer:

Increasing the radius of the circular motion or decreasing the speed of the object would both increase the period of uniform circular motion. This is due to the direct proportionality of period to radius and inverse proportionality to speed.

Explanation:

To determine which changes would increase the period of an object's uniform circular motion, let's recall the relationship between period (T), speed (v), and radius (r) of the circle. According to the formula for calculating the centripetal acceleration of an object in uniform circular motion (a = v²/r or a = 4π²r/T²), we can deduce that the period T is directly proportional to the radius and inversely proportional to the speed. Therefore, increasing the radius (Option I) would increase the period since T is proportional to the square root of the radius when speed is constant. Decreasing the speed of the object (Option IV) would also increase the period because the period is inversely proportional to speed.

The correct answer to the question of which changes would increase the period of motion in uniform circular motion are: I. Increase the radius of the circular motion and IV. Decrease the speed of the object. Options II and III would decrease the period, not increase it.