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Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

6 0

3 years ago

A major league baseball has a mass of 0.145 kg.

galben [10]

The momentum of the ball when it hits the ground is 4.89 kg.m/s.

The given parameters;

mass of the baseball, m = 0.145 kg
height of fall of the ball, h = 58 m

The final velocity of the ball when it hits the ground is calculated as follows;

$v_f ^2 = v_0^2 + 2gh\\\\v_f^2 = 0 + 2gh\\\\v_f^2 = 2gh\\\\v_f = \sqrt{2gh} \\\\v_f = \sqrt{2\times 9.8 \times 58}\\\\v_f = 33.72 \ m/s$

The momentum of the ball when it hits the ground is calculated as follows;

P = mv

P = 0.145 x 33.72

P = 4.89 kg.m/s

Thus, the momentum of the ball when it hits the ground is 4.89 kg.m/s.

Learn more here:brainly.com/question/22035809

6 0

2 years ago

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe

Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet = 327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

= 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

= 0.00353 × 128 × (327-40)

= 0.00353 × 128 × 287

= 129.68 J

Quantity of heat required to melt the bullet:

$Q = mH_f$

Q = 0.00353 × 24500

= 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

= 216.17 J

3 0

2 years ago

A 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the o

IgorLugansk [536]

Answer:

Explanation:

Using conservation of momentum

$m_1=59+220=279kg$

$m_2=2.1kg$

$U_1=12m/s$

$U_2=-44m/s$

$m_1U_1+m_2U_2=(m_1+m_2)V\\\\\frac{278\times 12-2.1\times 44}{279+2.1}=V\\\\V=11.58m/s$

6 0

2 years ago

Which of the following is the same in all frames of reference?

Ksju [112]

The correct option is B.
The length of an object, the mass of an object and the rate of time passage for an object can change depending on the situation which the object is subject to. For instance in space, the mass and the velocity of an object usually change. But, the value of the speed of light in the space is the same for all observers regardless of the motion of an object, that is, the speed of light is a constant.

4 0

3 years ago