What is the acceleration of an object that has a mass of 10kg and is pushed with a force of 50n

A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field

vazorg [7]

The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time

The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:

φ = BA

B = magnetic field strength, A = loop area

The area of the loop A is given by:

A = πr²

r = loop radius

Make a substitution:

φ = B2πr²

Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

Δφ = ΔB2πr²

ΔB = change in magnetic field strength

Make another substitution:

ℰ = |ΔB2πr²/Δt|

Given values:

ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s

Plug in and solve for ℰ:

ℰ = |(-0.20)(2π)(0.50)²/2.5|

ℰ = 0.13V

3 0

3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

$v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}$

$v_{y}=u Sin 30 - 2u Sin 30$

vy = - u Sin 30

The magnitude of final velocity is given by

$v = \sqrt{v_{x}^{2}+v_{y}^{2}}$

$v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}$

v = u

Thus, the velocity is same as it just reaches the ground.

6 0

3 years ago

Do you think the inner planets should be explored or should the money be spent on other things? Justify your opinion.

frosja888 [35]

Answer:

yes i think that inner planets should be explored because if we ever find new life or are able to live there then i think its worth takeing a risk.

Explanation:

4 0

3 years ago

An object spins in a horizontal circle with a radius of 15.0 cm. The rotations are timed and the amount of time it takes for it

bezimeni [28]

Answer:

1.7 m/s

Explanation:

Relevant Data provided as per the question below:-

Radius = 15.0 cm

Time = 0.56 s.

Based on the above information

The computation of the speed of the object is shown below:-

$Velocity = \frac{2\times \pi \times Radius}{Time}$

$Velocity = \frac{2\times \frac{22}{7} \times 0.15}{0.56}$

$= \frac{0.942857}{0.56}$

= 1.683 m/s

or

= 1.7 m/s

Therefore for computing the speed of the object or velocity we simply applied the above formula by considering the pi and all other given data

6 0

3 years ago

I have two questions

Anettt [7]

Answer:

1)The energy transported by a wave is directly proportional to the square of the amplitude. So whatever change occurs in the amplitude, the square of that effect impacts the energy. This means that a doubling of the amplitude results in a quadrupling of the energy

2)Just as wavelength and frequency are related to light, they are also related to energy. The shorter the wavelengths and higher the frequency corresponds with greater energy. So the longer the wavelengths and lower the frequency results in lower energy.

Explanation:

8 0

3 years ago

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