A rectangular loop with an area of 2 m2 is placed perpendicular to a uniform magnetic field of 1 Tesla. The field’s magnitude is increased to 6 Tesla in 4 seconds. The magnitude of the induced emf is equal to:

Answer:

You need the definition of acceleration (a=Vf-Vi/t) and 1 equation of linear motion (deltaX = Vi×t + 1/2×a×t^2). Since you know a is constant (gravity) and you know your initial Vi to be 40 m/s and your final velocity Vf to be zero (maximum height), then you can use thhe definition of acceleration to find time.

-9.81m/s^2 = (0-40m/s)/t

t = (-40)/(-9.81) s

t = 4.077s

Now that you have time, you should know all but deltaX in the equation of linear motion.

dX = (40m/s)(4.077s) + (1/2)(-9.81m/s^2)(4.077s)^2

dX = (163.099m) — (81.549m)

dX = 81.55m

Acceleration can be calculated by dividing the change in velocity by the change in time. If we have a mass and force that makes the acceleration, the divide the force by mass to get the acceleration.

Acceleration is a physical quantity which describes the rate of change in velocity. Like velocity, acceleration is a vector quantity having both magnitude and direction.

If both speed and direction are changing and  if we want to know the acceleration at any particular moment in time, calculating acceleration can be challenging.

However, since just speed is changing, it is quite simple to determine average acceleration over time. After then, acceleration is equal to the time change divided by the velocity change.

a = ΔV/Δt

Similarly if have the force which accelerates a body of mass m, then the acceleration is the ratio of that force to the mass.

a = F/m

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Answer:

What are the choices?

Explanation:

Answer: they move side to side

Explanation:

i took the test

Explanation:

Formula for speed = Distance covered/Time

Therefore,

Option A is correct.

Answer:

a.distance÷timetaken

Explanation:

The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).

Answer:

The acceleration is 4

Explanation:

use the formula v= change in position/change in time for vecter quantity's which will be in word problems that say acceleration or velocity.

Answer:

grams= 4.8e-5

kilograms= 4.8e-8

centigrams= 0.0048

dekagrams= 4.8e-6

Explanation:

Answer:

line graph

Explanation:

Ik im late but i hope this helps lol

Answer:

80 hours :)))))))))))))

Answer:

True

Explanation:

Answer:

Hans Christian Oersted

Explanation:

Answer:

A. human society

Explanation:

Complete Question

A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed?

a) The final kinetic energy is one half of the initial kinetic energy.

b) The final kinetic energy is one third of the initial kinetic energy.

c) The final kinetic energy is one quarter of the initial kinetic energy.

d) The final kinetic energy is one ninth of the initial kinetic energy.

e) The final kinetic energy is equal to the initial kinetic energy

Answer:

The correct option is b

Explanation:

From the question we are told that

   The mass of the sled is  m  

   The  mass of the three birds is  [tex]m_b = 0.5 m[/tex]

Generally from the law of momentum conservation,

  The initial momentum of the boat  =  final  momentum of the boat

Hence

     [tex]m * v = (m_b+ m) * v_2[/tex]

=>  [tex]m v = (0.5m+ m) * v_2[/tex]

=>  [tex]m v = 1.5m* v_2[/tex]

Here v is the velocity of the boat before the birds landed and [tex]v_2[/tex] is the velocity after the birds landed

So

       [tex]v_2 = \frac{2}{3} v[/tex]

Generally the initial  velocity of the boat before the birds landed is mathematically represented as

      [tex]K_i = \frac{1}{2} * m * v ^2[/tex]

Generally the final  velocity of the boat when the birds have landed is mathematically represented as

      [tex]K_f = \frac{1}{2} * (m_b + m ) * v_2^2[/tex]

=>   [tex]K_f = \frac{1}{2} * (0.5m + m ) * (\frac{2}{3} v)^2[/tex]

=>  [tex]K_f = \frac{1}{2} * 1.5m * (\frac{2}{3} v)^2[/tex]

=>  [tex]K_f = \frac{1}{2} * 1.5m * \frac{4}{9} v^2[/tex]

=>  [tex]K_f = \frac{1}{2}m v^2 * \frac{1}{3}[/tex]

Comparing this equation with that of the initial  kinetic energy

=>  [tex]K_f = \frac{1}{3} K_i[/tex]      

The kinetic energy of the sled and birds is one-third the initial kinetic energy of the sled before the birds landed.

When an object is in motion, then the energy possessed by the object at the state of motion is known as its kinetic energy.

Given data -

The combined mass of three birds is, M = 0.5m.

Let m be the mass of the sled. Then applying the conservation of linear momentum as,

The initial momentum of the boat  =  final  momentum of the boat

mv = (M + m)v'

v is the velocity of the boat before the birds landed and v' is the velocity after the birds landed.

mv = (0.5m + m)v'

mv = 1.5mv'

v = 3/2 v'

or

v' =2/3v

The initial kinetic energy of the boat before the birds landed is mathematically represented as

[tex]K_{1}= \dfrac{1}{2}mv^{2}[/tex]

And the final kinetic energy of the boat when the birds have landed is mathematically represented as,

[tex]K_{2}=\dfrac{1}{2}(M+m)v'^{2}\\\\K_{2}=\dfrac{1}{2}(1.5m) \times 4/9v^{2}\\\\K_{2}=\dfrac{1}{3} \times \dfrac{1}{2}mv^{2}\\\\K_{2}=\dfrac{1}{3} \times K_{1}[/tex]

Thus, we can conclude that the kinetic energy of the sled and birds is one-third the initial kinetic energy of the sled before the birds landed.

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Answer:

cardiovascular endurance

Explanation:

In order for Tom to successfully complete the marathon, he should focus on being able to maintain a constant pace without tiring easily.

Answer:

10 kg

Explanation:

Answer:

x = 3472 [m]

Explanation:

To solve this problem we must use the following expression of kinematics:

[tex]v_{f}^{2}= v_{i}^{2}+(2*a*x)\\[/tex]

where:

Vi = initial velocity = 0

Vf = final velocity = 300 [km/h]

a = acceleration = 1 [m/s^2]

x = takeoff distance [m]

Note: The initial velocity is equal to zero as avion starts its movement from rest.

Now we have to convert units of kilometers per hour to meters per second, the final velocity.

[tex]300 [\frac{km}{h}]*[\frac{1h}{3600s} ]*\frac{1000m}{1km} \\= 83.33 [\frac{m}{s} ][/tex]

(83.33)^2 = 0 + (2*1*x)

2*x = 6943.88

x = 3472 [m]

Answer:

Thermal energy and electrical energy

Explanation:

Answer:

Hello!!! Princess Sakura here ^^

Explanation:

The two kinds of energy are thermal and electrical.

Answer:

18 m/s

Explanation:

Range of a projectile on level ground is:

R = v₀² sin(2θ) / g

14.3 m = v₀² sin(2×13°) / 9.8 m/s²

v₀ = 17.9 m/s

Rounded to two significant figures, the launch speed was 18 m/s.

If the bullet is launched at an angle of 50 degrees above ground level, the target will be struck. The angle remains the same. The launch angle obtained is 50 degrees.

Given:

The initial shot was fired at an angle of 50 degrees above ground.

The projectile's starting velocity (v) and magnitude of velocity will remain constant if it is shot at the same pace.

Let the angle of the projectile is x,

The horizontal component of velocity can be calculated as follows:

[tex]v(x) = v * cos(x)[/tex]

We can write:

since the horizontal part of velocity remains constant:

[tex]v(x1) = v(x2)[/tex]

[tex]cos(50) = cos(x)[/tex]

[tex]50 = x[/tex]

Therefore, if the projectile is launched at the same speed at a 50° angle above ground level, it will strike the target.

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The correct response is given when the angle is asked a question.

Answer:  Option A is your answer

Explanation:

Answer:

128 is the ans cuz N is also lnown as mass

Explanation:

128

Answer:

17kN

Explanation:

The force of 10 kN in the cable and the load of 10 kN both have a component along the crane arm.

The force F exerted by the crane arm on the pulley is the sum of these 2 components.

Component of force in cable along the crane arm = 10 cos 30°

Component of load along the crane arm = 10 cos 30°

Considering the forces on the pulley,

F = 10 cos 30° + 10 cos 30°

Force F = 17.32 kN = 17 kN

Answer:

Density

Explanation:

The definition of density of a substance is the quotient between its mass and its volume. Such results in the mass per unit of volume.

Answer:u

Laa             c

Explanation:

Answer:

The magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I is [tex]\pi[/tex] or 3.14

Explanation:

The magnetic field at center of a circular loop is given by

[tex]B = \frac{\mu _{o}I }{2R}[/tex]

Where B is the magnetic field

[tex]\mu _{o}[/tex] is the free space permeability constant ( [tex]\mu _{o}[/tex] = 4π × 10⁻⁷ N/A²)

[tex]I[/tex] is the current

and [tex]R[/tex] is the radius

For the magnetic field of a long straight wire, it is given by

[tex]B = \frac{\mu _{o}I }{2\pi R}[/tex]

Where B is the magnetic field

[tex]\mu _{o}[/tex] is the free space permeability constant ( [tex]\mu _{o}[/tex] = 4π × 10⁻⁷ N/A²)

[tex]I[/tex] is the current

and [tex]R[/tex] is the distance from the wire

Then, to calculate the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I, that will be

[tex]\frac{\mu _{o}I }{2R} \div \frac{\mu _{o}I }{2\pi R}[/tex]

= [tex]\frac{\mu _{o}I }{2R} \times \frac{2\pi R }{\mu _{o}I}[/tex]

= [tex]\pi[/tex]

(NOTE: [tex]\pi[/tex] = 3.14)

Hence, the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I is [tex]\pi[/tex] or 3.14.

Answer:

Height ball A will deflect the target is 2.65 m

Height ball A will deflectthe target is 2.2 m

Ball A, will deflect the target to greater height, thus i will throw ball A

Explanation:

Given;

mass of the target, m₁ = 5.00 kg

mass of the ball, m₂ = 1.5 kg

initial velocity of each throw, u = 12 m/s

Throwing ball A; apply the principle of conservation linear momentum for elastic collision;

m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂

The initial velocity of the target, u₁ = 0

The ball bounced back at the same speed, v₂ = -12 m/s

the velocity of the target after collision, = v₁

0 + 1.5 x 12 = v₁(5) + (-12 x 1.5)

18 = 5v₁ - 18

18 + 18 = 5v₁

36 = 5v₁

v₁ = 36/5

v₁ = 7.2 m/s

The vertical height reached by the target is given by;

v₁² = u₁² + 2gh

v₁² = 0 + 2gh

v₁² = 2gh

h = v₁² / 2g

h = (7.2)² / (2 x 9.8)

h = 2.65 m

Throwing ball B; apply the principle of conservation energy for the inelastic collision;

the kinetic energy of the ball will be converted to the potential energy of the target.

¹/₂m₁u₂² = mgh

¹/₂(1.5)(12)² = (5 x 9.8)h

108 = 49h

h = 108 / 49

h = 2.2 m

Ball A will deflect  the target to greater height, thus i will throw ball A.

Answer:

a. (18,8)

b. (3, 6)

c. (10, 10, 10)

d. (a+d, b+e, c+f)

Give me brainliest for this... Thanks

Answer:

should be d because friction allows things to go faster or slower