A 10 kg plank 3 meters in length extends off the edge of a pirate ship so that only 0.5 m remains on the deck. This is held in place by a 200 kg crate of rum sitting on top of the end of the plank on the deck. A 70 kg pirate is being forced to walk the plank. How far does he get from the deck before the plank tips and he falls?

Answer:

A) 1.1 m/s/s

Explanation:

There exist two forces on the object such that

[tex]F_1[/tex] = 65 N directed 59° clockwise from the positive x-axis

[tex]F_2[/tex] = 35 N at 32° clockwise from the positive y-axis

now we have

[tex]F_1 = 65 cos59\hat i - 65 sin59 \hat j[/tex]

[tex]F_2 = 35 sin32\hat i + 35 cos32 \hat j[/tex]

now the net force on the object is given as

[tex]F_{net} = F_1 + F_2[/tex]

[tex]F_{net} = (65 cos59 + 35 sin32)\hat i + (35cos32 - 65 sin59)\hat j[/tex]

[tex]F_{net} = 52\hat i - 26 \hat j[/tex]

so it's magnitude is given as

[tex]F_{net} = \sqrt{52^2 + 26^2} = 58.15 N[/tex]

now from Newton's II law we have

F = ma

[tex]a = \frac{58.15}{55} = 1.1 m/s^2[/tex]

To find the magnitude of the object's acceleration, resolve the forces into components and use Newton's second law and the Pythagorean theorem.

To find the magnitude of the object's acceleration, we need to first resolve the two forces into their x and y components. For the 65 N force, the x-component is 65 N × cos(59°) and the y-component is 65 N × sin(59°). For the 35 N force, the x-component is 35 N × sin(32°) and the y-component is 35 N × cos(32°).

Next, we add the x-components and the y-components separately to get the net force in the x and y directions. Then we use Newton's second law, F = ma, to find the acceleration in each direction. Finally, we use the Pythagorean theorem to find the magnitude of the acceleration.

By following these steps, we can calculate that the magnitude of the object's acceleration is 1.5 m/s², so the correct answer is C) 1.5 m/s².

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Answer:

60.4 J

Explanation:

The work done by the gas is given by:

[tex]W=p(V_f-V_i)[/tex]

where

p is the gas pressure

[tex]V_f[/tex] is the final volume of the gas

[tex]V_i[/tex] is the initial volume

We must convert all the quantities into SI units:

[tex]p=860 torr \cdot \frac{1.013\cdot 10^5 pa}{760 torr}=1.146\cdot 10^5 Pa[/tex]

[tex]V_i = 127 mL = 0.127 L = 0.127 dm^3 = 0.127 \cdot 10^{-3}m^2[/tex]

[tex]V_f = 654 mL = 0.654 L = 0.654 dm^3 = 0.654 \cdot 10^{-3}m^2[/tex]

So the work done is

[tex]W=(1.146\cdot 10^5 Pa)(0.654\cdot 10^{-3} m^3-0.127\cdot 10^{-3} m^3)=60.4 J[/tex]

Answer:

[tex]W=60.4 J[/tex]

Work done is 60.4 Joules (J)

Explanation:

Work done 'W' is given by:

[tex]W=P\triangle V[/tex]

Where:

ΔV is the change in Volume.

P is the pressure.

Change in Volume=Final Volume-Initial Volume

Initial Volume= 127 mL

Final Volume= 654 mL

ΔV=654-127 (mL)

ΔV=527 mL

ΔV=0.527 L

Pressure Conversion:

1 atm=760 torr

[tex]P=\frac{860}{760} \\P=1.1315\ atm[/tex]

Now,

[tex]W=P\triangle V[/tex]

[tex]W=1.1315\ atm*0.527\ L[/tex]

[tex]W=0.5963 L.atm[/tex]

In joule (J): (Conversion 1 atm. L=101.325 J)

[tex]W=0.5963\ L.atm*\frac{101.325 J}{1\ L.atm}[/tex]

[tex]W=60.4 J[/tex]

Work done is 60.4 Joules (J)

Answer:

True.

Explanation:

This statament is true.

Answer:

4.72 N

Explanation:

The charge density across each plate is given by:

[tex]\sigma = \frac{Q}{A}[/tex]

where

[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate

[tex]A=0.3 m^2[/tex] is the area of each plate

Solving,

[tex]sigma = \frac{5\cdot 10^{-6}C}{0.3 m^2}=1.67\cdot 10^{-5} C/m^2[/tex]

The force exerted by one plate on the other is given by:

[tex]F=\frac{Q\sigma}{2\epsilon_0}[/tex]

where

[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate

[tex]\sigma=1.67\cdot 10^{-5} C/m^2[/tex] is the surface charge density

[tex]\epsilon_0[/tex] is the vacuum permittivity

Substituting,

[tex]F=\frac{(5\cdot 10^{-6} C)(1.67\cdot 10^{-5} C/m^2)}{2(8.85\cdot 10^{-12}F/m)}=4.72 N[/tex]

We can find the force exerted by one plate of a parallel-plate capacitor on the other by utilizing the known quantities and the formula F = Q² / (A * ε), which arises from force definition and electric field for a parallel-plate capacitor.

The magnitude of the force between two charges is given by Coulomb's law: F = k * (Q1 * Q2) / d², where k is the Coulomb's constant, Q1 and Q2 are the charges, and d is the distance. The electric field for a parallel-plate capacitor is E = Q / (A * ε), where ε is the permittivity of free space.

The force exerted on each plate of the capacitor is F = QE which, once substituted, becomes F = Q² / (A * ε) when charge, plate area, and permittivity are known. Given the values of the problem, by substituting into the equation, you can calculate the force between the plates.

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Answer:

0.8 m

Explanation:

For "6 N" force :

F = magnitude of the force = 6 N

r = moment arm = 0.4 m

Torque due to "6 N" force is given as

τ = r F

τ = (0.4) (6)

τ = 2.4 Nm

For " 15 N" force :

F' = magnitude of the force = 15 N

r' = moment arm = ?

τ' = Torque = 5 τ = 5 x 2.4 = 12 Nm

Torque due to "15 N" force is given as

τ' = r' F'

12 = r' (15)

r' = 0.8 m

So the moment arm for "15 N" force is 0.8 m

Answer:

2.17 s

Explanation:

Here, w0 = 33.3 rpm = 33.3 /60 rps

= 2 × pi × 33.3 / 60 rad/ s

= 3.4854 rad / s

w = 78 rpm = 78 / 60 rps

= 2 × pi × 78 / 60 = 8.164 rad / s

a = 2.15 rad/s^2

Use first equation of motion for rotational motion

w = w0 + a t

8.164 - 3.485 = 2.15 × t

t = 2.17 s

D. The chemical potential energy of real battery decreases, it develops internal resistance and therefore the potential difference across its terminals decreases if its current increases.

Answer:

The cannon recoils with a force of 332.5 N

Explanation:

By Newton's third law Recoil force on cannon = Force in shell.

Force in shell = Mass of shell x Acceleration of shell

Mass of shell = 3.5 kg

Acceleration of shell = 95 m/s²

Force in shell = 3.5 x 95 = 332.5 N

Recoil force on cannon = 332.5 N

So, the cannon recoils with a force of 332.5 N

[tex]\boxed{P=96.09MW}[/tex]

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

[tex]R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega[/tex]

So we can calculate the power loss as:

[tex]P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}[/tex]

Finally, the power loss due to resistance in the line is 96.09MW

Answer:

82 pF

Explanation:

diameter = 8.2 cm, distance, d = 2.1 mm = 0.0021 m,

dielectric constant, k = 3.7

radius = 1/2 x diameter = 8.2 / 2 = 4.1 cm = 0.041 m

The formula for the capacitance of parallel plate capacitor is given by

[tex]C = \frac{k \varepsilon _{0}\times A}{d}[/tex]

[tex]C = \frac{3.7 \times 8.854\times 10^{-12} \times \pi \times 0.041 \times 0.041}{0.0021}[/tex]

C = 8.23 x 10 ^-11 F

C = 82 pF

Answer:

B) Its velocity is perpendicular to the acceleration.

Explanation:

For general projectile motion, the horizontal acceleration is 0 and the vertical acceleration is -g.  This is true for all points on the trajectory.

At the highest point, the vertical velocity is 0.  So you have only a horizontal velocity as well as a vertical acceleration.  So the two are perpendicular.

(a) 1) The field is in the direction of the electron's initial velocity

The electric field is in a direction opposite to the initial velocity of the electron.

Let's remind that, when an electric charge is immersed in an electric field:

- if the charge is positive, the charge experiences a force in the same direction as the electric field direction

- if the charge is negative, the charge experiences a force in the opposite direction to the electric field direction

In this case, we have an electron: so the electric force exerted on the electron will be in a direction opposite to the direction of the electric field. Since the electron is accelerated in a direction opposite to the electron's initial velocity, this means that the electric force is in a direction opposite to the initial velocity, and so the electric field must be in the same direction as the electron's initial velocity.

(b) [tex]4.13\cdot 10^{-4} m[/tex]

We have:

Electron's initial velocity: [tex]u=5.25\cdot 10^6 m/s[/tex]

Electric field magnitude: [tex]E=1.9 \cdot 10^5 N/C[/tex]

Electron charge: [tex]q=-1.6\cdot 10^{-19} C[/tex]

Mass of the electron: [tex]m=9.11\cdot 10^{-31}kg[/tex]

The electric force exerted on the electron is:

[tex]F=qE=(-1.6\cdot 10^{-19} C)(1.9\cdot 10^5 N/C)=-3.04\cdot 10^{-14}N[/tex] (the negative sign means the direction of the force is opposite to its initial velocity)

The electron's acceleration is given by:

[tex]a=\frac{F}{m}=\frac{3.04\cdot 10^{-14} N}{9.11\cdot 10^{-31} kg}=-3.34\cdot 10^{16} m/s^2[/tex]

Now we can use the SUVAT equation:

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

where

v = 0 is the final speed (the electron comes to rest)

d is the total distance travelled by the electron

Solving for d,

[tex]d=\frac{v^2-u^2}{2a}=\frac{0-(5.25\cdot 10^6 m/s)^2}{2(-3.34\cdot 10^{16} m/s^2)}=4.13\cdot 10^{-4} m[/tex]

(c) [tex]1.57\cdot 10^{-10}s[/tex]

We can use the following equation:

[tex]a=\frac{v-u}{t}[/tex]

where we have

[tex]a=-3.34\cdot 10^{16}m/s^2[/tex] is the electron's acceleration

v = 0 is its final speed

[tex]u=5.25\cdot 10^6 m/s[/tex] is the initial speed

t is the time it takes for the electron to come at rest

Solving for t,

[tex]t=\frac{v-u}{a}=\frac{0-(5.25\cdot 10^6 m/s)}{-3.34\cdot 10^{16} m/s^2}=1.57\cdot 10^{-10}s[/tex]

(d) [tex]5.25\cdot 10^6 m/s[/tex]

This part of the problem is symmetrical to the previous part. In fact, the force exerted on the electron is the same as before (in magnitude), but in the opposite direction. This also means that the acceleration is the same (in magnitude), but in the opposite direction.

So we have:

u = 0 is the initial speed of the electron

[tex]a=3.34\cdot 10^{16}m/s^2[/tex]

[tex]d=4.13\cdot 10^{-4} m[/tex] is the distance covered to go back

So we can use the following equation:

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

to find v, the new final speed:

[tex]v=\sqrt{u^2 +2ad}=\sqrt{0^2 + 2(3.34\cdot 10^{16} m/s^2)(4.13\cdot 10^{-4} m)}=5.25\cdot 10^6 m/s[/tex]

Answer:

320 mm

Explanation:

The equivalent length for a minor loss is:

Le = K D / f

where K is the loss coefficient, D is the diameter, and f is the friction factor.

Le = (0.4) (20 mm) / 0.025

Le = 320 mm

The equivalent length (Le) for the elbow with a loss coefficient (K) of 0.4, a pipe with a friction factor of 0.025, and a diameter of 20 mm, is 320 mm.

To calculate the equivalent length (Le) for the elbow in terms of pipe diameter, we use the relationship between the loss coefficient (K), the friction factor (f), and the diameter (D) of the pipe, given by the equation Le = (K * D) / f. In this case, the loss coefficient for the elbow is 0.4, the friction factor is 0.025, and the diameter of the pipe is 20 mm. Plugging these values into the equation, we get:

Le = (0.4 * 20 mm) / 0.025 = 320 mm.

Therefore, the equivalent length for the elbow is 320 mm.

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

      19792.5+11.5m = 32416.5

        m = 1097.97 kg

Mass of car = 1098 kg

Answer:1074.26

Explanation:just got it right on my accelus

Answer:

0.96 m/s

Explanation:

Consider the motion of the ball before collision with the block

h = height from which the ball is dropped = length of the wire = 1.01 m

m = mass of the ball = 1.67 kg

v = speed of the ball just before collision with the block = ?

Using conservation of energy for the ball

Kinetic energy of the ball at the bottom = Potential energy of the ball at the top

(0.5) m v² = mgh

(0.5) v² = gh

(0.5) v² = (9.8) (1.01)

v = 4.45 m/s

consider the collision between the ball and the block :

m = mass of the ball = 1.67 kg

v = velocity of the ball just before collision with the block = 4.45 m/s

v' = velocity of the ball just after collision with the block

M = mass of the block = 2.59 kg

V = velocity of the block just before collision with the ball = 0 m/s

V' = velocity of the block just after collision with the ball

Using conservation of momentum

mv + MV = mv' + MV'

(1.67) (4.45) + (2.59) (0) = 1.67 v' + (2.59) V'

7.43 = 1.67 v' + (2.59) V'

[tex]V' = \frac{(7.43 - 1.67 v')}{2.59}[/tex]                                eq-1

Using conservation of kinetic energy

(0.5) mv² + (0.5) MV² = (0.5) mv'² + (0.5) MV'²

mv² + MV² = mv'² + MV'²

(1.67) (4.45)² + (2.59) (0)² = 1.67 v'² + (2.59) V'²

using eq-1

(1.67) (4.45)²  = 1.67 v'² + (2.59) ((7.43 - 1.67 v')/2.59)²

v' = - 0.96 m/s

the negative sign indicates the direction which is opposite to its direction before colliding with the block.

To find the velocity of the ball just after the elastic collision, apply the formula for one-dimensional elastic collision. Given that the ball initially fell from a certain height, calculate its initial velocity using the conservation of energy principle. Insert these results, along with the given masses, into the elastic collision formula to achieve the answer.

This question explores the principles of conservation of momentum and the properties of an elastic collision. Given that this collision is perfectly elastic, we can use the formula for an elastic collision in one-dimension:

v1' = ((m1 - m2) / (m1 + m2)) * v1 + ((2 * m2) / (m1 + m2)) * v2

where v1' is the velocity of the ball after collision, m1 and m2 are the masses of the ball and the block respectively, v1 and v2 are the velocities of the ball and block before the collision. The ball was initially dropped from a height, so we'd first need to calculate its velocity just before collision using the conservation of energy principle:

v1 = sqrt(2* g * h)

where g is the acceleration due to gravity and h is the height (length of the wire). After figuring out the value of v1, we just need to plug the values into the first equation to find v1' - velocity of the ball after collision.

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Answer:

4 m/s2 in negative acceleration

Explanation:

If An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds, it is 4 m/s2 in negative acceleration.

Hope this helps!

Answer:

[tex]-4\frac{m}{s^2}[/tex]

Explanation:

The object changes its speed over some time, this means that there is an acceleration.

It has a uniformly accelerated movement.

The Formula for finding speed in a uniformly accelerated motion is

[tex]a=\frac{V_{f}-V_{o}}{t}[/tex]

[tex]V_{o}= 65\frac{m}{s}\\V_{f}= 25\frac{m}{s} \\t= 10s[/tex]

Replace

[tex]a=\frac{(25-65)\frac{m}{s} }{10 s}\\ a=\frac{-40\frac{m}{s} }{10s}\\a= -4\frac{m}{s^2}[/tex]

Acceleration gives us a negative value this means that it is slowing.

Answer:

167.85 rev / s

Explanation:

r = 12 cm = 0.12 m, m = 3 x 10^-16 kg, F = 4 x 10^-11 N

F = m r w^2

where, w is the angular velocity.

4 x 10^-11 = 3 x 10^-16 x 0.12 x w^2

w = 1054.1 rad / s

w = 2 π f

f = w / 2 π = 1054.1 / (2 x 3.14) = 167.85 rev / s

The number of revolutions given by the calculated frequency value in which the centrifuge would be operated is 167.8 Hz.

Recall :

We calculate the angular velocity, ω thus :

ω² = F/mr

ω² = [tex] \frac{4 \times 10^{-11}}{3 \times 10^{-16} \times 0.12 = 11.11 \times 10^{5}[/tex]

ω = [tex] \sqrt{1.11 \times 10^{6}} = 1053.56 rad/s[/tex]

Frequency = 1053.56 ÷ (2π)

Frequency = 167.68 Hz

Therefore, the Number of revolutions per seconds would be about 167.8 Hz

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Answer:

Angle of ray makes with the vertical is 62.1 degree

Explanation:

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

[tex]tan\theta_i = \frac{0.7}{0.98}[/tex]

[tex]\theta_i = 35.5^0[/tex]

now by Snell'a law at that interface we have

[tex]\mu_1 sin\theta_i = \mu_2 sin \theta_r[/tex]

now we will have

[tex]1.52  sin35.5 = 1.40 sin\theta_r[/tex]

[tex]\theta_r = 39.12^0[/tex]

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

[tex]\mu_2 sin\theta_i' = \mu_{air} sin\theta[/tex]

[tex]1.40 sin39.12 = 1 sin\theta[/tex]

[tex]\theta = 62.1^0[/tex]

The problem involves applying Snell's law twice, once for the brine-to-oil interface and then for the oil-to-air interface, with angles calculated from the normal lines. Severally, the use of the Pythagorean theorem gives the angle in brine, Snell's law then gives the angle in oil, and finally in the air. The desired angle is found as an augmentation to 90 degrees since it should be calculated from the axis, not the normal line.

Firstly we need to calculate the angle that the ray makes with the norm or perpendicular line to the brine-oil interface. This can be done using the Pythagorean theorem, given that we have the height of the brine (0.98m), and the horizontal distance from the axis to the point of intersection (0.7m). The angle is then the arcsin of the opposite side (0.7m) over the hypotenuse (√[(0.7m)2+(0.98m)2]) in radians. (Remember that inverse sin is used to calculate angles). Let's call this angle θ1.

Using Snell's law (n1sinθ1 = n2sinθ2), where the indices of refraction n1 and n2 are for brine (1.52) and oil (1.40) respectively, we can figure out θ2, the angle that the ray makes with the normal in the oil part. The key detail is that angles are always measured from the normal line, which is perpendicular to the interfaces.

Finally, similar calculations will grant what is to be found - the angle the ray makes with the vertical axis in the air (with the index of refraction n=1). This is θ3 calculated from the equation n2sinθ2 = n3sinθ3, where θ2 is the angle that we just found and n2 is the index of refraction for oil. You need to remember that Snell's law applies for any two mediums with a clear border in between through which the light passes.

Finally, the desired angle between the ray's line in the air and the vertical axis will be 90° - θ3 (since θ3 was an angle with the normal line, which is defined as perpendicular to the axis).

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Answer: [tex]1.1052g/cm^{3}[/tex]

Explanation:

Density [tex]D[/tex] is a characteristic property of a material and is defined as the relationship between the mass [tex]m[/tex] and volume [tex]V[/tex] of a specific substance or material. So, the density of the asteroid is given by the following equation:

[tex]D=\frac{m}{V}[/tex]   (1)

On the other hand, we know the asteroid has a mass [tex]m=1000kg[/tex] and is spherical. This means its volume is given by the following formula:

[tex]V=\frac{4}{3}}\pi r^{3}[/tex]   (2)

Where [tex]r=\frac{d}{2}=\frac{1.2m}{2}=0.6m[/tex]  is the radius of the sphere and is half its diameter [tex]d[/tex].

Knowing this, we can calculate the volume:

[tex]V=\frac{4}{3}}\pi (0.6m)^{3}[/tex]   (3)

[tex]V=0.904m^{3}[/tex]   (4)

Substituting (4) in (1):

[tex]D=\frac{1000kg}{0.904m^{3}}=1105.242\frac{kg}{m^{3}}[/tex]   (5) This is the density of the asteroid, but we were asked to find it in [tex]\frac{g}{cm^{3}}[/tex]. This means we have to make the conversion:

[tex]D=1105.242\frac{kg}{m^{3}}.\frac{1000g}{1kg}.\frac{1m^{3}}{(100cm)^{3}}[/tex]

Finally:

[tex]D=1.1052\frac{g}{cm^{3}}[/tex]

Answer:

the answer is that u need to divide the 26 by the 22 and then u add seven to that number and then u have to multiply by 2. And you will get your answer

Explanation:

Answer:

Object's acceleration, a = 0.15 m/s²

Explanation:

It is given that,

Net force acting on the object, F = 7.5 N

Mass of an object, m = 47 kg

We need to find the acceleration of the object. It can be calculated using second law of motion as :

F = ma

a = acceleration of the object

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

[tex]a=\dfrac{7.5\ N}{47\ kg}[/tex]  

[tex]a=0.15\ m/s^2[/tex]

So, the acceleration of the object is 0.15 m/s². Hence, this is the required solution.

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

[tex]f=mg\ sin\theta[/tex]

[tex]f=1100\ kg\times 9.8\ m/s^2\ sin(4)[/tex]

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

In essence, the force of friction that prevents the car from sliding down the incline is equal to the component of the weight of the car down the incline. After performing the necessary calculations, this is found to be roughly 766.9 Newtons.

In response to your question, the force of friction keeping the car from sliding down the incline can be calculated using basic physics principles. Firstly, we need to calculate the component of the gravitational force parallel to the incline, which is given by F_gravity = m * g * sin(theta), where m is the mass of the car, g is the acceleration due to gravity, and theta is the angle of incline.

Substituting the given values, we get F_gravity = 1100 kg * 9.8 m/s² * sin(4°), which approximates to 766.9 N.

Now, since the car is stationary and not sliding down, this suggests that the force of friction equals the component of the weight down the incline. Therefore, the force of friction keeping the car from sliding down is approximately 766.9 Newtons.

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Let [tex]\theta[/tex] be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

[tex]\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]

The current has velocity vector (relative to the Earth)

[tex]\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath[/tex]

The swimmer's resultant velocity (her velocity relative to the Earth) is then

[tex]\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}[/tex]

[tex]\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath[/tex]

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

[tex]\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ[/tex]

which is approximately 41º west of north.

The swimmer should swim approximately 36.87 degrees in the north-west direction to counteract the eastward ocean current and reach Victoria, British Columbia directly from Port Angeles, Washington. This is derived by considering the velocity vectors of the swimmer and the current.

This question is about the interaction of velocity vectors which is a concept from physics. The swimmer's speed and the ocean current each represent a vector with both a magnitude (speed) and direction. We must factor in these two vectors to work out where the swimmer should aim to swim.

Assuming the north direction as +y axis and east as +x axis, the swimmer intends to swim north with velocity 4.0 km/h and the water's current is moving towards east with velocity 3.0 km/h. In order for the swimmer to move directly north, the swimmer needs to swim against the current, which means westwards. By utilizing the Pythagorean Theorem and trigonometric principles, we can calculate the angle of her direction which is arctan(3/4) in the north-west direction or around 36.87 degrees from north.

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Answer:

Wavelength: the distance between two crests of an electromagnetic wave

Frequency: the rate at which electromagnetic waves oscillate

Explanation:

For a transverse wave (such as the electromagnetic waves), we can define the following quantities:

Wavelength: it corresponds to the distance between two consecutive crests (or between two consecutive troughs) of the wave

Frequency: the number of complete oscillations of the wave per unit time

Period: it is the time it takes for the wave to complete one oscillation - it is equal to the reciprocal of the frequency

Amplitude: it is the maximum displacement of the wave, measured with respect to its equilibrium position

Answer:

0.0018 V

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the particle is equal to the electric potential energy lost:

[tex]\frac{1}{2}mv^2 = q\Delta V[/tex]

where

[tex]m=3.8\cdot 10^{-6} kg[/tex] is the mass of the particle

[tex]v=36 m/s[/tex] is the final speed of the particle

q = -2.7 C is the charge

[tex]\Delta V[/tex] is the potential difference between the two points

Solving for [tex]\Delta V[/tex], we find

[tex]\Delta V= \frac{mv^2}{q}=\frac{(3.8 \cdot 10^{-6} kg)(36 m/s)^2}{-2.7 C}=-0.0018 V[/tex]

The particle has been accelerated by this potential difference: since it is a negative charge, it means that the particle has moved from a point at lower potential towards a point of higher potential.

So, since the initial point is A and the final point is B, the result is

[tex]V_B - V_A = 0.0018 V[/tex]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

[tex]u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s[/tex]

[tex]v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s[/tex]

Now we find the centripetal acceleration which is given by

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

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

[tex]a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2[/tex]

we also have a tangential acceleration, which is given by

[tex]a_t = \frac{v-u}{t}[/tex]

where

t = 17.0 s

Substituting values,

[tex]a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2[/tex]

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

[tex]a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2[/tex]

Answer:

a = 1.406 m/s²

Explanation:

We are told that, the speed of the train decreases from 94.0 km/h to 46.0 km/h

Let's convert both to m/s.

Thus,

v1 =94 km/h =(94 x 10)/36 =26.11 m/s

v2=46 km/h =(46 x 10)/36=12.78 m/s

The formula to calculate the tangential acceleration is given by;

a_t = -dv/dt

Where;

dv is change in velocity

dt is time difference

dv is calculated as; dv = v1 - v2

Thus, dv = 26.11 - 12.78 = 13.33 m/s

We are given that, t = 17 seconds

Thus;

a_t = -13.33/17 = -0.784 m/s²

The negative sign implies that the acceleration is inwards.

Now, let's calculate the radial acceleration;

a_r = v²/r

Where;

r is the radius of the path = 140m

v is the velocity at the instant given

a_r is radial acceleration.

Thus,

a_r = 12.78²/140 = 1.167 m/s²

Now, the tangential and radial components of acceleration are perpendicular to each other. Thus, we can use using Pythagoreas theorem to find the resultant acceleration;

Thus;

a² = (a_t)² + (a_r)²

Plugging in the relevant values, we have;

a² = (-0.784)² + (1.167)²

a² = 1.976545

a = √1.976545

a = 1.406 m/s²

Answer:

7.8 V

Explanation:

When resistors are connected in series, they are connected in the same branch of the circuit. This means that the current flowing each resistor is the same, while the sum of the voltage drops across each resistor is equal to the potential difference of the power supply:

[tex]V= V_1 + V_2 + V_3[/tex]

In this circuit we have:

V = 15 V

[tex]V_1 = 4.1 V[/tex]

[tex]V_2 = 3.1 V[/tex]

So, the voltage drop across resistor 3 is

[tex]V_3 = V-V_1 - V_2 = 15 V - 4.1 V - 3.1 V=7.8 V[/tex]

Final answer:

The exact potential drop across the third resistor, when three resistors are connected in series to a 15-V power supply and the voltage drops across the first two are 4.1V and 3.1V, is 7.8 volts.

Explanation:

The question asks for the exact potential drop across resistor 3 when three resistors are connected in series across a 15-V power supply, and the potential drops across the first two resistors are known. In a series circuit, the total voltage supplied is equal to the sum of individual voltage drops across all components. Given that the voltage drops across resistors 1 and 2 are 4.1 volts and 3.1 volts respectively, the voltage drop across resistor 3 can be calculated by subtracting the sum of the voltage drops across resistors 1 and 2 from the total voltage supplied by the power source.

Therefore, the calculation is:

Total voltage supplied by the power source = 15 volts

Sum of voltage drops across resistor 1 and 2 = 4.1 volts + 3.1 volts = 7.2 volts

Voltage drop across resistor 3 = Total voltage - Sum of voltage drops across resistor 1 and 2 = 15 volts - 7.2 volts = 7.8 volts.

Answer:

(e) 98,1 KJ

Explanation:

The engine produces 19%; it means, it rejects 81% of energy. ⇒ 81/19=4.26 times.

The engine produces 23 kJ; it means it rejects 23 * 4.26 = 98.05263 kJ

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

[tex]L_2 - L_1 = 10 Log \frac{I_2}{I_1}[/tex]

so here we need to find the ratio of two intensity

it is given as

[tex]Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}[/tex]

[tex]Log\frac{I_2}{I_1} = \frac{60 - 30}{10}[/tex]

[tex]Log\frac{I_2}{I_1} = 3[/tex]

now we have

[tex]\frac{I_2}{I_1} = 10^3[/tex]

so it is

d) 1000 times

Final answer:

A baseball pitched horizontally at 101.0 mi/h will travel 60.5 feet to home plate in about 0.408 seconds; during that time, it will fall approximately 2.68 feet due to gravity.

Explanation:

To calculate how far a baseball would fall vertically when thrown horizontally at 101.0 mi/h, we can divide the problem into two separate motions - horizontal and vertical. Since the vertical and horizontal motions are independent, we can analyze them separately. First, we need to convert the speed to feet per second (1 mi/h = 1.467 ft/s), multiply 101.0 mi/h by 1.467 to get the horizontal velocity in feet per second, and then use that to find the time it takes for the ball to travel the horizontal distance to home plate.

Horizontal distance to home plate = 60.5 ft. Horizontal speed Vh = 101.0 mi/h * 1.467 ft/s/mi/h = 148.137 ft/s. Time to reach home plate t = distance/speed = 60.5 ft / 148.137 ft/s = 0.408 s (approximately).

Next, to find out how far the ball falls vertically, we'll use the formula for the distance an object falls due to gravity: d = 0.5 * g * t2, where g is the acceleration due to gravity. In standard English units, g is approximately 32.2 ft/s2. Plugging in the numbers, we get d = 0.5 * 32.2 ft/s2 * (0.408 s)2 = 2.68 ft (approximately).

Thus, a baseball thrown horizontally at 101.0 mi/h will fall about 2.68 feet by the time it reaches home plate.

The correct answer is that the ball would fall vertically by approximately 2.63 ft by the time it reached home plate, 60.5 ft away.

To solve this problem, we can use the kinematic equation for the vertical motion of the ball under the influence of gravity, assuming no air resistance. The equation is:

[tex]\[ y = v_{0y} t + \frac{1}{2} g t^2 \][/tex]

where:

- y  is the vertical displacement,

- [tex]\( v_{0y} \)[/tex] is the initial vertical velocity (which is 0 m/s in this case since the ball is thrown horizontally),

-  g  is the acceleration due to gravity (9.81 m/s, but we will use 32.2 ft/s for consistency with the imperial units given in the problem),

-  t is the time of flight.

Since the initial vertical velocity [tex]\( v_{0y} \)[/tex] is 0, the equation simplifies to:

[tex]\[ y = \frac{1}{2} g t^2 \][/tex]

We need to find the time  t  it takes for the ball to travel 60.5 ft horizontally. The horizontal velocity [tex]\( v_x \)[/tex] is given as 101 mi/h. We convert this to feet per second:

[tex]\[ v_x = 101 \text{ mi/h} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \][/tex]

[tex]\[ v_x = 101 \times 5280 \times \frac{1}{3600} \text{ ft/s} \][/tex]

[tex]\[ v_x = 147.67 \text{ ft/s} \][/tex]

Now we can calculate the time  t it takes to travel 60.5 ft:

[tex]\[ t = \frac{\text{distance}}{\text{velocity}} = \frac{60.5 \text{ ft}}{147.67 \text{ ft/s}} \][/tex]

[tex]\[ t \ = 0.409 \text{ s} \][/tex]

Using this time, we can find the vertical displacement y :

[tex]\[ y = \frac{1}{2} \times 32.2 \text{ ft/s}^2 \times (0.409 \text{ s})^2 \][/tex]

[tex]\[ y = \frac{1}{2} \times 32.2 \times 0.167 \text{ ft} \][/tex]

[tex]\[ y = 16.7 \times 0.167 \text{ ft} \][/tex]

[tex]\[ y \ = 2.63 \text{ ft} \][/tex]