In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. What is the magnitude of the impulse delivered by the bat to the ball

Answers

Answer 1

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns


Related Questions

on heating gases expand ______________ than solids and liquids

Answers

Answer:

On heating gases expands than solid and liquid because the molecules gets seperated and the volume increases that causes to expand.

Explanation:

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a bicycle has a momenum of 36kg•m/s and a velocity of 4m/s. what is the mass of the bicycle?

Answers

p=mv, so the mass is 9 kg.

The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the speed of the wave?

Answers

Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

A piano has a mass of 99 kg. What is the weight of the piano?
A. 1030 N
B. 842 N
C. 1129 N
D. 970 N
Help!!

Answers

Answer:

[tex]D.\ 970\ N[/tex]

Explanation:

Given:

Mass of Piano: 99kg

Required:

Calculate its weight

The weight of an object is calculate as thus;

[tex]Weight = Mass\ (m) * Acceleration\ due\ to\ gravity\ (g)[/tex]

[tex]m = 99kg[/tex] and [tex]g = 9.8m/s^2[/tex]

The formula becomes

[tex]Weight = 99kg * 9.8m/s^2[/tex]

[tex]Weight = 970.2\ kgm/s^2[/tex]

[tex]Weight = 970.2\ N[/tex]

[tex]Weight = 970\ N[/tex] Approximated

Hence, the weight of the piano is 970N

PLEASE HELP ASAP. IT'S URGENT

Answers

Answer:

Q1 acceleration = 16m/s²

Q2 Net force = 9N North

Explanation:

Q1 Using the formula F=ma

Q2 R = F1 + F2

7.Why are we able to drink cold drink with straw from a bottle?

Answers

Answer:

BETTER EXPLANATION

Explanation:

When you drink from a straw, you create a little space of low pressure inside your mouth and in the top of the straw. Then the air outside the straw pushes down on the surface of the drink and forces the liquid up through the straw and into your mouth.

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What far vertically will it travel before hiting the ground A. 40 m B. 30 m C. 60 m D. 50 m

Answers

Answer:

First, let's think in the vertical problem:

The acceleration will be the gravitational acceleration:

g = 9.8 m/s^2

a = -9.8 m/s^2

For the velocity, we integrate over time:

v(t) = (-9.8 m/s^2)*t + v0

Where v0 is the initial velocity, in this case  v0 = 30m/s.

v(t) =  (-9.8 m/s^2)*t + 30m/s

Now, for the position we integrate again over time, and get:

P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0

Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m

p(t) = (-4.9m/s^2)*t^2 + 30m/s*t

Now, the maximum vertical height is reached when:

v(t) = 0m/s = -9.8m/s^2*t + 30m/s

t = 30m/s/9.8m/s^2 = 3.06s

Now we can evaluate the vertical position in t = 3.06s

p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m

So, rounding down, the correct option is: C. 60 m

3.) [15 points] A physics teacher is on the west side of a small lake and wants to swim across and up at a point directly across from his starting point. He notices that there is a current in the lake and
that a leaf floating by him travels 4.2m [S] In 5.0s. He is able to swim 1.9 m/s in calm water,
(a) What direction will he have to swim in order to arrive at a point directly across from his position?​

Answers

Answer:

The teacher should swim in a direction 29.24° North of East

Explanation:

Given that the there is a water  current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;

The speed of the water current is given by the speed of the floating leaf traveling with the water current  

Distance traveled by the leaf = 4.2 m South

Time of travel of the leaf = 5.0 s

Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current

Swimming peed of the teacher, v = 1.9 m/s

To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.

Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim

Therefore;

1.9 × sin(θ) = 0.84

sin(θ) = 0.84/1.9 = 0.44

θ = 26.24°

That is the teacher should swim in a direction 29.24° North of East.

To cross the lake the teacher has to swim in a direction 29.24° North of the East

Finding the direction of speed required:

The speed of the water current can be derived from the speed of the floating leaf :

The distance traveled by the leaf L = 4.2 m South

Time taken T = 5s

So, the speed of the leaf is:

u = 4.2/5

u = 0.84 m/s South

So, the speed of the current is 0.84 m/s South

Now, it is given that the speed of the teacher is, v = 1.9 m/s East

To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.

Let, the speed of the teacher makes an angle of θ from the EAST.

So, the North component is given by:

v(north) = vsinθ

it must be equal to the speed of the current:

vsinθ = u

1.9 × sinθ = 0.84

sinθ = 0.84/1.9

sinθ = 0.44

θ = 26.24°

The teacher should swim in a direction 29.24° North of East.

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A train travels 120 km at a speed of 60 km/h, makes a stop for 0.5 h, and then travels the next 180 km at a speed of 90 km/h. What is the average speed of the train for this trip? 9th grade level pls FFFFFFFFFFFAAAAAAAAAAASSSSSSSSSSSSSTTTTTTTTTTTTT

Answers

Answer:

average speed =  66.67  km/h

Explanation:

In order to find the average speed of the train, you need to calculate the total distance traveled, divided by the time it took to cover that distance. So for the total distance:

Distance= 120 km + 180 km = 300 km

For the total time we need to add three different quantities, two of which we need to derived based on the information provided:

time for first part of the trip:

[tex]time_1=\frac{D_1}{v_1} =\frac{120}{60} \,h= 2\,h[/tex]

for the time of the stop:

[tex]time_2=0.5\,\,h[/tex]

for the last part of the trip:

[tex]time_3=\frac{180}{90} \,h= 2 \,\,h[/tex]

Which gives a total of 4.5 hours

Then, the average speed is:   300/4.5   km/h = 66.67  km/h

In the school cafeteria, a trouble-making child blows a 12.0 g spitball through a 25.0 cm straw. The force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw, followed by a constant force through the rest of the straw. The force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally.
(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.
(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?
(c) What is the maximum force F0, that acts on the spitball?

Answers

Answer:

A.) Find the attached file

B.) 1.536 J

C.) 123N

Explanation:

Given that the force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw,

Half of the straw = 12.5 cm

Let Force F = dependent variable

And distance x = independent variable

A.) Find the attached file for the graph.

Sinnce the force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally

B.) Given that the mass M = 12g = 12/1000 = 0.012kg

And Velocity V = 16m/s

The workdone = the kinetic energy of the split ball

WD = 1/2mv^2

Substitutes m and V into the formula

WD = 1/2 × 0.012 × 16^2

WD = 1.536 Joule

C.) The maximum force acts on the spit ball will be at maximum kinetic energy.

Since work done = force × distance

Where distance = 12.5 cm

F × S = 1/2mv^2

Substitutes all the parameters

F × 0.0125 = 1.536

Make F the subject of formula

F = 1.536/0.0125

F = 122.88 N

The maximum force F0, that acts on the spitball is therefore equal to 123 N approximately.

We have that for the Question "(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.

(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?

(c) What is the maximum force F0, that acts on the spitball?"

It can be said that

a) The sketch is attachedb) The workdone  = [tex]1.536J[/tex]c) Maximum force = [tex]6.4N[/tex]

From the question we are told

mass of spitball = 12g through a 25.0 cm straw

Generally the equation for workdone is mathematically given as

[tex]= \frac{1}{2}mv^2 - 0\\\\= \frac{1}{2} * 12*10^{-3} * 16^2\\\\= 1.536J[/tex]

Maximum force

[tex]w = \frac{1}{2}*\frac{x_0}{2}*\frac{F_0}{2} + x_0*\frac{F_0}{2}\\\\w = \frac{5}{8}x_0F_0[/tex]

Therefore,

[tex]F_0 = \frac{8w}{5x_0}\\\\F_0 = \frac{8*1.536}{5*25*10^[-2]}\\\\F_0 = 6.4N[/tex]

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a) A conductor carrying a current I = 12.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A magnetic force per unit length of 0.110 N/m acts on the conductor in the negative y direction. Determine the magnitude of the magnetic field in the region through which the current passes.

Answers

Answer:

8.8 mT

Explanation:

Current through the conductor = 12.5 A

Magnetic force per unit length on the wire = 0.110 N/m

Recall that the magnetic force per unit length on a current carrying conductor is in a uniform magnetic field is

[tex]\frac{F}{l}= IBsin\alpha[/tex]

where B is the magnetic field magnitude

[tex]I[/tex] is the current in the conductor

α is the angle the conductor makes with the  magnetic field ( since it is perpendicular in this case, α is 90°)

imputing values into the equation, we'll have

0.11 = 12.5 x B x sin 90°

but sin 90° = 1, therefore, we have

0.11 = 12.5B

B = 0.11/12 = 8.8 x 10^-3 T

or rather = 8.8 mT

A monkey is sitting at the top of a tree 25 m high from the ground level. A person standing on the ground wants to feed the monkey. He uses a bow and arrow to launch the food to the monkey. If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the food, how should the person aim the arrow containing the food? Group of answer choices

Answers

Answer:

He should aim it at the monkey

Explanation:

The question is related to projectile motion of an arrow from a bow such that the arrow is able to reach a specific target

Therefore, since the monkey is going to drop from the tree the instant the person launches the food, then for the monkey to be able reach the food, we have;

Final level position of the launched food should be ≥ The level of the monkey

As the monkey is dropping from the tree the destination of the launched food should be at the monkey so that before the food gets to the position where the monkey was when launching the food with the arrow, the monkey will be slightly lower so as to be able to catch the food or return up the tree to get the food rather than the food falling to the ground.

Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?

Answers

Answer:

According to super hero logic , nothing will happen to him.

But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.

He may / may not be affected but his suit will be damaged for sure as it is made of metal.

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please help me with this ​

Answers

Answer:

12 Ω

Explanation:

Data obtained from the question include:

Resistor 1 (R1) = 50 Ω.

Resistor 2 (R1) = 30 Ω.

Resistor 3 (R3) = 20 Ω.

Resistor 4 (R4) = R

Galvanometer reading = Zero deflection.

The Resistor 4 (R4) in the Wheatstone bridge can be obtained as follow:

Since the galvanometer gives zero deflection, it means the bridge is balanced. Therefore,

R1/R2 = R3/R4

50/30 = 20/R

Cross multiply

50 x R = 30 x 20

Divide both side by 50

R = (30 x 20)/50

R = 12 Ω

Therefore, the value of R in the wheatstone bridge is 12 Ω.

Stopping distance of vehicles When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v0) and the braking capacity, or deceleration that is caused by braking. A car travelling at speed 72km/hr suddenly applies the brake with the deceleration of 5m/s2. Find the stopping distance of the car.

Answers

Answer:

Stopping distance = 40m

Explanation:

Given the following :

Initial speed of vehicle before applying brakes = 72km/hr

Converting km/hr to m/s:

72km/hr = [(72 * 1000)m] / (60 * 60)

72km/hr = 72,000m / 3600s

72km/hr = 20m/s

Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2

From the 3rd equation of motion:

v^2 = u^2 + 2as

Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance

Final velocity when the car stops will be 0

Therefore ;

v^2 = u^2 + 2as

0 = 20^2 + 2(-5)(s)

0 = 400 - 10s

10s = 400

s = 400/10

s = 40m

Therefore, the stopping distance of the car = 40 meters

calculate the energy dissipated by an electric iron which draws a current of 5A from a240v power supply for 25minutes​

Answers

Explanation:

Given,

I = 5 A

V = 240 V

T = 25 mins = 1500 sec

Now,

Energy dissipated = IVT= 5×240×1500 = 1800000 J

What are the signs of the velocity, force, and acceleration (components) afterthe cart is released and is moving qway from the motion sensor? (The positive xdirection is toward the motion sensor, to the right.

Answers

Answer:

car is moving away its direction is negative.

the speed must also be negative.

speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

Explanation:

In the exercise they indicate that the direction to the motion sensor is positive, as they indicate that the car is moving away its direction is negative.

The speed of the car is

           v = (x₂-x₁) / t

As the positions are negative, and the car moves away the speed must also be negative.

The analysis for acceleration must be very careful if the speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

a 6 letter word a way of explaining an object or event using a set of facts

Answers

Explanation:

A theory is a way of explaining an object or event using a set of facts.

A car travels 18 km / h to the northwest. What is the speed of this car seen from another car going south at 60 km / h?

Answers

Answer:

73.8 km/h

Explanation:

The velocity of object A relative to object B is:

v = va − vb

Let's say east is +x and north is +y, and angles are measured from +x.

The velocity of the first car in the x direction is:

18 km/h cos 135° = -9√2 km/h

The velocity of the first car in the y direction is:

18 km/h sin 135° = 9√2 km/h

The velocity of the second car in the x direction is:

60 km/h cos 270° = 0 km/h

The velocity of the second car in the y direction is:

60 km/h sin 270° = -60 km/h

The relative velocity in the x direction is:

vₓ = -9√2 km/h − 0 km/h = -9√2 km/h ≈ -12.7 km/h

The relative velocity in the y direction is:

vᵧ = 9√2 km/h − (-60 km/h) = 9√2+60 km/h ≈ 72.7 km/h

The speed, or magnitude of the velocity, is:

v = √(vₓ² + vᵧ²)

v = √((-12.7 km/h)² + (72.7 km/h)²)

v ≈ 73.8 km/h

Select the correct answer.
According to the Universal Law of Gravitation, every object attracts every other object in the universe. Why can’t you feel the force of attraction between you and Mars?
A.
There is no force of attraction between you and Mars.
B.
Your mass is too low.
C.
Mars is a larger planet than Earth.
D.
Mars is a long distance away.

Answers

Answer:

D. Mars is a long distance away

A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of aluminum is 2698.4 kg / m3

Answers

Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]

We need to find the thrust and the force. The mass of the liquid displaced is given by :

[tex]m=dV[/tex]

V is volume

Weight of the displaced liquid

W = mg

[tex]W=dVg[/tex]

So,

[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]

So, the thrust and the force is 1562.53 N.

Two trains run in the opposite direction with speeds of v1 = 15 m / s and v2 = 20 m / s. A passenger on the first train (the one on v1) notes that train 2 takes 6 s to pass on its side. What is the length of the second train? (The passenger is supposed to be immobile looking through the window)

Answers

Answer:

210 m

Explanation:

The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.

It takes 6 seconds for the train to pass, so the length of the train is:

(35 m/s) (6 s) = 210 m

A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air has an index of refraction of 1.00. At what minimum angle will light in the glass tube be totally reflected at the glass/air interface?

Answers

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

         sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

        θ = sin⁻¹ n₂ / n₁

let's calculate

       θ = sin⁻¹ (1.00 / 1.50)

       θ = 41.8º

The following passage has not been edited. There is an error in each line. Write the

incorrect word and the correction in your answer sheet against the correct question

number. The first one has been done as an example. ( 1 x 4 = 4 )

Community service sensitize people to Error: sensitize ; Correction: sensitizes

other‟s needs and supports inclusive (a) Error: _______ ; Correction: ______

development to the underprivileged (b) Error: _______ ; Correction: ______

sections with society. Courses about social (c) Error: ______ ; Correction: _______

work prepares frontline workers to (d) Error: _______ ; Correction: ______​

Answers

Answer:

(a) Error: Other's ; Correction: Others'

(b) Error: to ; Correction: for

(c) Error: with ; Correction: of

(d) Error: prepares ; Correction: prepare

Explanation:

a)  The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.

b)  The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.

c)  The error is in the word "with"  and the correct word will be "of" as of indicates relationships between other words including things that made of other things.

d)  The error is in the word "preapres" and the correct word will be "prepare".

Radar uses radio waves of a wavelength of 2.5 m . The time interval for one radiation pulse is 100 times larger than the time of one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect

Answers

Answer:

The minimum distance to the object that the radar can detect is 124.995 m

Explanation:

Here, we are to calculate the shortest distance to an object the radar in the question can detect.

Mathematically;

v = c/λ

Where v is the frequency, c is the speed of light and λ is the wave length

Thus;

v = (3 * 10^8)/2.5 = 1.2 * 10^8 Hz

Mathematically, the time period

t = 1/v = 1/(1.2 * 10^8) = 0.000000008333 = 8.333 * 10^-9 sec

From the question, we are told that the transmitting time is 100 times a single oscillation

Transmitting time = 100 * one oscillation

Hence Transmitting time = 100 * 8.33 * 10^-9 = 8.33 * 10-7

Mathematically;

Minimum distance =( Transmitting time * speed of light)/2 =

(8.33 * 10^-7 * 3 * 10^8)/2 = 124.995 m

Let us treat a helicopter rotor blade as a long thin
rod, as shown in Fig. 8–49. (a) If each of the three rotor
helicopter blades is 3.75 m long and has a mass of 135 kg,
calculate the moment of inertia of the three rotor blades
about the axis of rotation. (b) How much torque must the
motor apply to bring the blades from rest up to a speed
of 6.0 rev/s in 8.0 s?
Rotor

Answers

Answer:

(a) 1900 kg m²

(b) 8950 Nm

Explanation:

(a) The moment of inertia of a rod about its end is I = ⅓mL².

For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:

I = 3 (⅓ (135 kg) (3.75 m)²)

I = 1900 kg m²

(b) Net torque = moment of inertia × angular acceleration

∑τ = Iα

First, find the angular acceleration.

ω₀ = 0 rad/s

ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s

t = 8.0 s

α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²

∑τ = Iα

∑τ = (1900 kg m²) (4.71 rad/s²)

∑τ = 8950 kg m² / s²

∑τ = 8950 Nm

On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the average speed for the entire journey is 40km/h?

Answers

Answer:

v = 46.67 km/h

Explanation:

We will use the following formula throughout this numerical:

s = vt

where,

s = distance covered

v = speed

t =  time taken

FOR FIRST 30 km:

s = 30 km

v = 30 km/h

t = t₃₀ = ?

Therefore,

30 km = (30 km/h)(t₃₀)

t₃₀ = (30 km)/(30 km/h)

t₃₀ = 1 h

FOR TOTAL 100 km:

s = 100 km

v = 40 km/h (Average Speed)

t = total time = ?

Therefore,

100 km = (40 km/h)(t)

t = (100 km)/(40 km/h)

t = 2.5 h

FOR LAST 70 km:

s = 70 km

t₇₀ = t - t₃₀ = 2.5 h - 1 h = 1.5 h

v = v₇₀ = ?

Therefore,

70 km = v(1.5 h)

v = 70 km/1.5 h

v = 46.67 km/h

A square loop 24.0 cm on a side has a resistance of 6.10Ω. It is initially in a 0.665-T magnetic field, with its plane perpendicular to magnetic field B but is removed from the field in 40.0ms. Calculate the electric energy dissipated in this process.

Answers

Answer:

[tex]E=6.01\times 10^{-3}\ J[/tex]

Explanation:

It is given that,

Side of a square loop, l = 24 cm = 0.24 m

Resistance of loop, R = 6.1 ohms

Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms

We need to find the electrical energy dissipated in this process.

Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :

[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]

If I is induced current then,

[tex]V=IR[/tex]

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]

Power is given by voltage times current. So,

[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]

Now, energy is given by the product of power and time. So,

[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]

Now putting all the values in above formula. So,

[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]

So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.

The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".

Magnetic field

According to the question,

Square loop's side, l = 24 cm or,

                                  = 0.24 m

Loop's resistance, R = 6.1 ohms

Initial magnetic field = 0.665 T

Final magnetic field = 0

We know the relation,

→ V = - [tex]\frac{dB}{dt}[/tex]

      = [tex]\frac{BA}{t}[/tex]

Also we know,

Current, V = IR

               I = [tex]\frac{V}{R}[/tex]

                 = [tex]\frac{BA}{tR}[/tex]

Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,

                          = [tex]\frac{(BA)^2}{tR}[/tex]

By substituting the values,

                          = [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]

                          = 6.01 × 10⁻³ J

Thus the response above is correct.                                  

Find out more information about magnetic field here:

https://brainly.com/question/26257705

A car drives at a constant speed of 21 m/s around a circle of radius 100 m.
What is the centripetal acceleration of the car?
O A. 4.8 m/s2
O B. 0.21 m/s2
O C. 3.1 m/s2
O D. 4.4 m/s2

Answers

Answer:

Option D. 4.4 m/s²

Explanation:

The following data were obtained from the question:

Velocity (v) = 21 m/s

Radius (r) = 100 m

Centripetal acceleration (a) =.?

The centripetal acceleration of the car can be obtained as follow:

Centripetal acceleration (a) = Velocity square (v²) / radius (r)

a = v²/r

a = 21²/100

a = 441/100

a = 4.41 ≈ 4.4 m/s²

Therefore, the centripetal acceleration of the car is 4.4 m/s².

PLEASE HELP ME ASAP. IT'S VERY IMPORTANT

Answers

Answer:

1) a.  52.41 m/s

b. The skier will be going 15.35 m/s slower

2)  103.68 m

3) 35,127 J

4) a.  88.825 kJ

(b) 16.36 %

5) 3,071.12 J

Explanation:

1) a. The given height of the hill, h = 140.0 m

The mass of the skier at the top of the hill, m = 85.0 kg

The acceleration due to gravity, g = 9.81 m/s²

The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J

From the principle of conservation of energy, we have;

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

m×g×h = 1/2×m×v²

116739 J = 1/2×85.0×v²

v² = 116739/(1/2*85.0)= 2746.8 m²/s²

v = √(2746.8 m²/s²) = 52.41 m/s

b. From 70 m up, we have;

The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

58,369.5 J = 1/2×85.0×v²

v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²

v = 37.06 m/s

The skier will be going 52.41 - 37.06 = 15.35 m/s slower

The skier will be going 15.35 m/s slower

2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load

The amount of work done by the electric hoist = 356,000 J

The mass of the load = 350.0 kg

The height to which the load is raised = h

The potential energy gained by the load = m×g×h = 350.0×9.81×h

356,000 J = 350.0×9.81×h

h = 356,000/(350.0*9.81) = 103.68 m

The height to which the load is lifted= 103.68 m

3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J

The final potential energy = 600*28.0*9.81= 164808 J

The velocity at point 3  = 4.5 m/s

The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J

The total energy at point 3 = 164808 + 6075 = 170,883 J

The energy loss = The initial potential energy at point 1 - Total energy at point 3

The energy loss = 206010 - 170,883 = 35,127 J

The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J

4) a. The heat energy absorbed = mass × specific heat capacity for water, [tex]C_{water}[/tex] × Temperature change

The mass of the water = 2.5×10² g = 0.25 kg

[tex]C_{water}[/tex] = 4,180 J/(kg·°C)

Initial temperature = 10.0°C

Final temperature = 95°C

The temperature change = 95.0°C - 10.0°C = 85.0°C

The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ

(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100

The heat supplied = 543 kJ

The efficiency = (88.825/543) × 100  = 16.36 %

5) The mass of the box = 115 kg

Force acting on the rope = 255 N

The angle of inclination of the force to the horizontal = 24.5°

The distance the box is displaced = 15.0 m to the right

The work done = Force applied × distance moved in the direction of the force

The work done = Force applied × distance moved in the direction of the force

Given that the load moves a distance 15.0 m to the right,we have;

The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) =  204.74 N

The work done = 204.7*15 = 3071.12 J

The amount of work done  = 3,071.12 J

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