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3 years ago

An object’s mass increases its

Physics

1 answer:

STatiana [176]3 years ago

4 0

Answer:

It is commonly known that, if you accelerate an object, its mass will increase; however, to understand why this phenomenon occurs, we mustn’t think of the object’s mass increasing. Instead, we should think of its energy. In physics, mass is just simply locked up energy. We call this type of mass, ‘inertial mass.’

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"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines a

sergey [27]

Answer:

B = 5.59x10⁹ T

Explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

$F = qvBsin(\theta)$

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

$B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T$

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!

7 0

3 years ago

Friction is defined as *

den301095 [7]

Answer:

O a force that opposes motion

5 0

3 years ago

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Aleonysh [2.5K]

The answer would be that they are close to water hope this helps!

6 0

3 years ago

Read 2 more answers

A circular loop of wire 78 mm in radius carries a current of 114 A. Find the (a) magnetic field strength and (b) energy density

Finger [1]

Explanation:

It is given that,

Radius of loop, r = 78 mm = 0.078 m

Current, I = 114 A

(a) Magnetic field strength of the circular loop is given by :

$B=\dfrac{\mu_oI}{2R}$

$B=\dfrac{4\pi \times 10^{-7}\times 114}{2\times 0.078}$

B = 0.000918 T

$B=9.18\times 10^{-4}\ T$

(b) Energy density at the center of the loop is given by :

$U=\dfrac{B^2}{2\mu_o}$

$U=\dfrac{(0.000918)^2}{2\times 4\pi \times 10^{-7}}$

$U=0.335\ J/m^3$

Hence, this is the required solution.

7 0

3 years ago

John throws a ball with a velocity of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

umka21 [38]

The horizontal component of the velocity is equal to: D. 15 m/s.

<u>Given the following data:</u>

Velocity = 30 m/s

Angle = 60°

To determine the horizontal component of the velocity:

The horizontal component of the velocity represents the influence of velocity in displacing an object or projectile in the horizontal direction.

Mathematically, the horizontal component of velocity is given by the formula:

$V_x = Vcos(\theta)$

Substituting the given parameters into the formula, we have;

$\\\\V_x = 30cos(60)\\\\V_x = 30 \times 0.5$

Horizontal component, Vx = 15 m/s

Read more on horizontal component here: brainly.com/question/24681896

6 0

3 years ago