On thanksgiving morning you take a turkey with mass and temperature into an oven and it stays there until the temperature of the turkey is . Determine the change in thermal energy for the turkey. The specific heat for turkey is .

Answers

Answer 1

Answer:

2.81 kJ/(kg-K).

Explanation:

The specific heat capacity of turkey is 2.81 kJ/(kg-K) which means that it takes 2.81 kJ of energy to heat one kg of turkey by increasing the temperature by one degree Kelvin. Specific heat capacity is defined as the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 kelvin of temperature so the specific heat capacity for turkey is 2.81 kJ/(kg-K).


Related Questions

If the depth of water in a well is 10 m, what is the pressure exerted by it on the
bottom of the well? (Use g = 10 m/s)
[Ans: 10 N/m]

Answers

Answer:

Let d be the density of the water  (1000 kg / m^3   eq to 1 gm / cm^3)

P = d g h     for the pressure due to a column at the bottom of the column.

P = 1000 kg / m^3 * 10 m/s^2 * 10 m = 10^5 kg / m * s^2 = 10^5 N/m

A pitching machine is programmed to pitch baseballs horizontally at a speed of 126 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. For each of the object pairings listed below, determine the correct relative speed.

a. The speed of the pitching machine relative to the truck
b. The speed of the pitched bell relative to the truck
c. The speed of the pitching machine relative to you
d. The speed of the pitched ball relative to you

Answers

Explanation:

a) zero, since the machine is mounted on the truck

b) 126 km/hr

c) 85 km/hr

d) 126 km/hr + 85 km/hr = 211 km/hr

A compact disc rotates at 500 rev/min. If the diameter of the disc is 120 mm, (a) What is the tangential speed of a point at the edge of the disc? (b) At a point halfway to the center of the disc?

Answers

Answer:

(a) the tangential speed of a point at the edge is 3.14 m/s

(b) At a point halfway to the center of the disc, tangential speed is 1.571 m/s

Explanation:

Given;

angular speed of the disc, ω = 500 rev/min

diameter of the disc, 120 mm

radius of the disc, r = 60 mm = 0.06 m

(a) the tangential speed of a point at the edge is calculated as follows;

[tex]\omega = 500 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} = 52.37 \ rad/s[/tex]

Tangential speed, v = ωr

                               v = 52.37 rad/s  x 0.06 m

                              v = 3.14 m/s

(b) at the edge of the disc, the distance of the point = radius of the disc

   at half-way to the center, the distance of the point = half the radius.

r₁ = ¹/₂r = 0.5 x 0.06 m = 0.03 m

The tangential velocity, v = ωr₁

                                       v = 52.37 rad/s x 0.03 m

                                       v = 1.571 m/s

Một ống dây điện thẳng dài có lõi sắt, tiết diện ngang của ống S = 20 cm2

, chiều dài
1 m, hệ số tự cảm L = 0,44 H. Cường độ từ trường trong ống dây là H = 0,8.103 A/m. Từ
thông gửi qua tiết diện ngang của ống bằng

3

0
1,6.10 Wb

. Cường độ dòng điện chạy

qua ống dây là

Answers

Answer:

sgsbssbduebubbeeifirjeirneejrbb8m!keoejr

d

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An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small appliances on her trip.

a. What is the ratio of turns in the primary and secondary coils of her transformer?
Np / Ns = ____________

b. What is the ratio of input to output current?
Iin /Iout = ___________

c. How could a Chinese person traveling in the United States use this same transformer to power her 220 V appliances from 120 V?

Answers

Answer:

(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Explanation:

Given;

input voltage, [tex]V_p[/tex] = 220 V

output voltage, [tex]V_s[/tex] = 120 V

General transformer equation is given as;

[tex]\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]

where;

Np is number of turns in the primary coil

Ns is number of turns in the secondary coil

Is - is the secondary current or output current

Ip - is the primary current or input current

(a) The ratio of turns in the primary and secondary coils of her transformer;

[tex]\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833[/tex]

(b) The ratio of input to output current;

[tex]\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55[/tex]

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)

A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.

At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.

Answers

Answer:

Explanation:

The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.

From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.

1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.

Answers

Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

 

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

[tex]I =\frac{q}{t}[/tex]  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]

Therefore,

t = 1.27 x 10⁹ s

A 6.90 kg block is at rest on a horizontal floor. If you push horizontally on the 6.90 kg block with a force of 12.0 N. It just
starts to move.
What is the coefficient of static friction?
Numeric Response

Answers

[tex]\mu = 0.177[/tex]

Explanation:

Let's look at the forces on the two axes:

[tex]x:\:\:\:F - f_n = F - \mu N = 0\:\:\:\:\:\;(1)[/tex]

[tex]y:\:\:\:N - mg = 0\:\:\:\:\:\:\:\:\:(2)[/tex]

Substituting (2) into (1) and solving for [tex]\mu[/tex], we get

[tex]F = \mu mg[/tex]

[tex]\mu = \dfrac{F}{mg} = \dfrac{12.0\:\text{N}}{(6.9\:\text{kg})(9.8\:\text{m/s}^2)} = 0.177[/tex]

A 1,071.628 N painter needs to climb d=1.926 m up a ladder (measured along its length from the point where the ladder contacting the ground), without the ladder slipping. The uniform ladder is 12.014 m long and weighs 250 N. It rests with one end on the ground and the other end against a perfectly smooth vertical wall. The ladder rises at an angle of theta=51.96 degrees above the horizontal floor. What is friction force in unit of N that the floor must exert on the ladder? Use g = 10 m/s2 if you need to .

Answers

The frictional force in unit of N that the floor must exert on the ladder is approximately 232.216 N

The known values are;

The weight of the painter = 1,071.628 N

The height to which the painter needs to climb along the ladder = 1.926 m

The length of the ladder = 12.014 m

The weight of the ladder = 250 N

The points where one of the ladder's ends is resting = On the ground

The points where the other end of the ladder is resting = A perfectly smooth wall

The angle with which the ladder rises above the horizontal floor = 51.96°

The acceleration due to gravity, g ≈ 10 m/s²

The unknown values include;

The friction force that the floor must exert on the ladder

The strategy to be used;

At equilibrium, the sum of moments about a point is zero

Finding the moments about the point of contact where the ladder rests on the wall, P, is given as follows;

At equilibrium, the sum of clockwise, [tex]M_{CW}[/tex], moment about P = The sum of the counterclockwise, [tex]M_{CCW}[/tex]moment about P

[tex]\mathbf{M_{CCW}}[/tex] = (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

[tex]\mathbf{M_{CW}}[/tex] = 12.014 × cos(51.96°) × [tex]\mathbf{F_N}[/tex]

Where;

[tex]\mathbf{F_N}[/tex] = The normal reaction of the of the ground on the end of the ladder that rests on the floor

[tex]\mathbf{M_{CCW}}[/tex] = [tex]\mathbf{M_{CW}}[/tex]

∴ (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250 = 12.014 × cos(51.96°) × [tex]F_N[/tex]

We get;

6,665.3068846 N·m =  7.40316448688 m × [tex]F_N[/tex]

[tex]\mathbf{F_N}[/tex] = 6,665.3068846 N·m/(7.40316448688 m) = 900.332135 N

The normal reaction of the floor on the ladder, [tex]\mathbf{F_N}[/tex] = 900.332135 N

Taking moment about the point the ladder rests on the floor, R, gives;

[tex]M_{CCW}[/tex] = 12.014 × sin(51.96°) × [tex]F_W[/tex]

Where;

[tex]\mathbf{F_W}[/tex] = The normal reaction at the wall

[tex]M_{CW}[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

At equilibrium, we have, [tex]M_{CCW}[/tex] = [tex]M_{CW}[/tex]

Therefore;

12.014 × sin(51.96°) × [tex]F_W[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

9.46199511627 m × [tex]F_W[/tex] = 2,197.22861125 N·m

[tex]F_W[/tex] = 2,197.22861125 N·m/(9.46199511627 m)

The reaction of the wall, [tex]\mathbf{F_W}[/tex] = 232.216206 N

We note that also at equilibrium, the sum horizontal forces = 0

The horizontal forces acting  on the ladder = The normal reaction on the, [tex]F_W[/tex] wall and the friction force on the ground, [tex]\mathbf{F_f}[/tex]

∴ At equilibrium; [tex]\mathbf{F_W}[/tex] + [tex]\mathbf{F_f}[/tex] = 0

[tex]\mathbf{F_f}[/tex] = -[tex]\mathbf{F_W}[/tex]

[tex]\mathbf{F_W}[/tex]  = 232.216206 N

Therefore;

The frictional force in unit of N that the floor must exert on the ladder, [tex]\mathbf{F_f}[/tex] = 232.216206 N 232.216 N.

(The coefficient of friction, μ = [tex]\mathbf{F_N}[/tex]/[tex]\mathbf{F_W}[/tex] = 900.332135/232.216206 ≈ 3.877).

Learn more about the force of friction here;

https://brainly.com/question/8859573

https://brainly.com/question/14111224

https://brainly.com/question/23567411

A toy rocket that weights 10 N blasts straight up from ground level with an initial kinetic energy of 40J. at the exact top of its trajectory its total mechanical energy is 140J. To what vertical height does it rise?

Answers

Answer:

14 m

Explanation:

Applying,

P.E = WH............ Equation 1

Where P.E = potential energy at the highest vertical height, W = weight of the toy rocket, H = Vertical height.

Note: At the highest vertical height, the kinetic energy is 0 J

Therefore,

Mechanical energy(M.E) = Potential Energy(P.E)

make H the subject of the equation

H = P.E/W......... Equation 2

From the question,

Given: W = 10 N,  P.E = 140 J

Substitute these values in equation 2

H = 140/10

H = 14 m

1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.

Answers

Answer:no

Explanation:because 0.9*(30*60)=0.9*1800=1620

The turtle has already won the race

Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

What will be the speed of the rabbit and the turtle?

It is given

[tex]V_{t} = 0.9 \frac{m}{s}[/tex]

[tex]V_{r} = 9 \frac{m}{s}[/tex]

[tex]D=1500 m[/tex]

Time taken by turtle  

 [tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]

[tex]T=1666 minutes= 27 hours[/tex]

Time taken by  rabbit

[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]

[tex]T=166 minutes[/tex]

since rabbit started 30 minutes after turtle then

[tex]T= 136+30=196 minutes[/tex]

[tex]T= 3.2 hours[/tex]

Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

To know more about average velocity follow

https://brainly.com/question/6504879

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N , how hard does the wall push on you

Answers

Answer:

44 N

Explanation:

Given that,

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.

It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.

What is the name of the invisible line that runs
down the center of the axial region?

Answers

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Explanation:

The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.

the momentum of a spring coil when the external compressing force is removed b the difference between the final momentum and the initial momentum of the object c the backward momentum felt by an object or person exerting force on another object d the difference between the total momentum of the system after impact and the total momentum of the system before impact

Answers

Answer:

the correct one is b

the difference between the final moment and the initial moment

Explanation:

The momentum is related to the moment

          I = ΔP

          ∫ F dt = p_f - p₀

where p_f and p₀ are the final and initial moments, respectively

When checking the different answers, the correct one is b

the difference between the final moment and the initial moment

An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown

Answers

Answer: [tex]283.2\times 10^{-9}\ nC[/tex]

Explanation:

Given

Cross-sectional area [tex]A=0.4\ cm^2[/tex]

Dielectric constant [tex]k=4[/tex]

Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]

Distance between capacitors [tex]d=5\ mm[/tex]

Maximum charge that can be stored before dielectric breakdown is given by

[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]

Answer:

The maximum charge is 7.08 x 10^-8 C.

Explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]

At what angle torque is half of the max

Answers

At what angle torque is half of max

find the exit angle relative to the horizontal in an isosceles triangle with 36 °​

Answers

what

what

what

what

sorry

sorrry

sorry

The position of the image obtained by convex lens when object is kept beyond 2F1(F: principal focus of the convex lens)
A. between F2 and 2F2
B. at 2F2
C. beyond 2F2
D. at infinity

Answers

Answer:

Between F2 and 2F2

Explanation:

Diagram attached from Teachoo.

Link to website if you need to refer

https://www.teachoo.com/10838/3118/Convex-Lens---Ray-diagram/category/Concepts/

Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.


52 cm4


72 cm4


32 cm4


24 cm4


2 cm4

Answers

Answer:

Minimum Area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum Area of rectangle

Computation:

Minimum Area of rectangle = Length of rectangle x Width of rectangle

Minimum Area of rectangle = 6 x 4

Minimum Area of rectangle = 24 centimeter²

A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?

Answers

Answer:

28 j

Explanation:

because when you add you get 28

3. The figure below shows the motion of a car. It starts from the origin, O travels 8m
towards the east and then 12m towards the west.
D
8m.
X
X-8
12m.w
()What is the net displacement D from the origin to the final position?
(ii) What is the total distance travelled by the car?

Answers

Answer:

i. -4m

ii. 20m

Explanation:

The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)

Total distance = 8m going east + 8m back to origin + 4m west = 20m

An evacuated tube uses an accelerating voltage of 55 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?

Answers

Answer:

v = 4.4 x 10⁷ m/s

Explanation:

The kinetic energy of the electrons will be equal to the energy supplied by the electric voltage:

Kinetic Energy = Electric Energy

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

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

e = charge on electron = 1.6 x 10⁻¹⁹ C

V =Voltage = 55 kV = 55000 V

Therefore,

[tex]\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(v)^2 = (1.6\ x\ 10^{-19}\ C)(55000\ V)\\\\v^2 = \frac{(2)(8.8\ x\ 10^{-16}\ J)}{9.1\ x\ 10^{-31}\ kg}\\\\v = \sqrt{19.34\ x\ 10^{14}\ m^2/s^2}[/tex]

v = 4.4 x 10⁷ m/s

Your forehead can withstand a force of about 6.0 kN before it fractures. Your cheekbone on the other hand can only handle about 1.3 kN before fracturing. If a 140 g baseball hits your head at 30.0 m/s and stops in 0.00150 s,

Required:
a. What is the magnitude of the ball's acceleration?
b. What is the magnitude of the force that stops the baseball?
c. What force does the baseball apply to your head? Explain?
d. Are you in danger of a fracture if the ball hits you in the forehead?

Answers

Answer:

Explanation:

a)

Final velocity v = 0 ; initial velocity u = 30 m/s , time t = .0015 s

v = u + a t

0 = 30 m/s + a x .0015 s

a = - 30 / .0015

= - 20000 m / s²

b )

Magnitude of force = m x a

= .140 kg x 20,000 m / s²

= 2800 N = 2.8 kN.

c )

The force applied by baseball = 2.8 kN .

d )

Since ball can withstand a force of 1.3 kN so it will break if 2.8 kN force acts on it . SO, head will fracture.

Why must scientists be careful when studying
nanotechnology?

Answers

Answer:

When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.

Hope it helps u:)

Use the image of Potential vs. position in 1D to match each scenario with subsequent motion.

A (+) charge is placed at A and can only move in the x-direction. When it is released, what will happen?
Correct answer:
It will move to the left
A (-) charge is placed at A and can only move in the x-direction. When it is released, what will happen?
Incorrect answer:
It remains at where it was placed.
A (-) charge is placed at B and can only move in the x-direction. When it is released, what will happen?
Correct answer:
It remains at where it was placed.
A (+) charge is placed at B and pushed slightly to the right; it can only move in the x-direction. What will happen?
Correct answer:
It will move to the right.
A (-) charge is placed at B and pushed slightly to the right; it can only move in the x-direction. What will happen?
Correct answer:
It will oscillate around B
Continuing the previous exercise, determine the nature of work (for each force listed, not net force), KE and PE for:


1. A + charge moving away from a + charge, from rest, under field force only.


KE

[ Select ]

0 PE

[ Select ]

0 Work

[ Select ]

0


2. A + charge moving away from a + charge, from rest, with applied force slowing it.


Work is

[ Select ]

0


3. A - charge moving toward a + charge under field force only.


KE

[ Select ]

0 PE

[ Select ]

0 Work is

[ Select ]

0


4. A - charge moving toward a + charge with applied force slowing it.


Work is

[ Select ]

0


5. An applied force pulls a negative charge away from a positive charge.


Work is

[ Select ]


6. An applied force pushes 2 like charges together.


Work is

[ Select ]

Answers

Answer:

incorporators and it is the one you for the delay to get it for now that the new to me to the same as last week to week in my opinion of your

What is inertia of motion?​

Answers

Explanation:

Inertia of motion

It is also known as Newton's first law of motion.

It states that,

An object remains in a state of rest or of uniform motion in a straight line unless compelled to change its state by an applied external force.

8. A boat moving initially at 6.5 km hr due southwest crosses a river that is flowing due south at 3 km hr.
What is the magnitude and direction of the boat relative to the ground? If the river is 1.5 mi wide how long
does it take the boat to cross?

Answers

Answer:

a)  v = 8,878 km / h, θ’= 238.8º,  b) t = 1890.9 s

Explanation:

a) In this exercise we must find the resulting speed of the boat.

Let's use trigonometry to break down the speed of the boat (v1)

            cos 225 = v₁ₓ / v₁

            sin 225 = v_{1y} / v₁

            v₁ₓ = v₁ soc 225

            v_{1y} = v₁ sin 225

            v₁ₓ = 6.5 cos 225 = -4.596 km / h

            v_{1y} = 6.5 sin 225 = -4.596 km / h

to find the velocity we add each component

           vₓ = v₁ₓ

           vₓ = - 4,596 km / h

           v_y = v_{1y} + v₂

           v_y = -4.596 - 3

           v_y = - 7,596 km / h

Now let's compose the speed

Let's use the Pythagorean theorem for the module

           v = [tex]\sqrt{v_x^2 + v_y^2 }[/tex]

           v = Ra 4.596² + 7.596²

           v = 8,878 km / h

Let's use trigonometry for the direction

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹  ( [tex]\frac{-7.596}{ -4.596}[/tex] )

          θ = 58.8º

measured from the positive side of the x-axis

          θ'= 180 + 58.8

          θ’= 238.8º

b) Let's reduce the river width to the SI system

          x = 1.5 miles (1,609 km / 1 mile) = 2,414 km

to cross the river the speed is on the x axis which is the width of the river

         v = x / t

         t = x / v

         t = 2.414 /4.596

         t = 0.525 h

let's reduce to the SI system

         t = 0.525 h (3600 s / 1h)

         t = 1890.9 s

Calculate the buoyant force due to the surrounding air on a man weighing 600 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3.

Answers

Answer:

[tex]F_b= 0.720 N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=600N[/tex]

Average density [tex]\rho=1.20kg/m^3[/tex]

Mass

[tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{600}{9.81}[/tex]

[tex]m=61.22kg[/tex]

Generally the equation for Volume is mathematically given by

[tex]V =\frac{ mass}{density}[/tex]

[tex]V= \frac{61.22}{1000}[/tex]

[tex]V=0.06122 m^3[/tex]

Therefore

Buoyant force [tex]F_b[/tex]

[tex]F_b=\rho*V*g[/tex]

[tex]F_b= rho_air*V*g[/tex]

[tex]F_b= 0.720 N[/tex]

An alternating voltage is connected in series with a resistance R and an inductance L If the potential drop across the
resistance is 200 V and across the inductance is 100V
then the applied voltage is
V 223.6
V 2006
V 300
V50
Please help me

Answers

Answer:

oh my God I got really confused right now

When placed 1.18 m apart, the force each exerts on the other is 11.2 N and is repulsive. What is the charge on each

Answers

Answer:

[tex]q=41.62\ \mu C[/tex]

Explanation:

Given that,

Force between two objects, F = 11.2 N

Distance between objects, d = 1.18 m

We need to find the charge on each objects. The force between charges is as follows :

[tex]F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C[/tex]

So, the charge on each sphere is [tex]41.62\ \mu C[/tex].

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