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2 years ago

A particle with mass 3×10−2 kg

Physics

1 answer:

liraira [26]2 years ago

8 0

The speed of the particle when it enters the magnetic-field region is 120,000 m/s.

<h3>Speed of the particle</h3>

Magnetic force on the particle is given as;

F = qvB

where;

q is magnitude of the charge
v is speed of the particle
B is magnetic field strength

From Newton's second law, force on an object is given as;

F = ma

where;

m is mass of the particle
a is acceleration

ma = qvB

v = ma/qB

v = (3 x 10⁻² x 12)/(3 x 10⁻⁶ x 1)

v = 120,000 m/s

Thus, the speed of the particle when it enters the magnetic-field region is 120,000 m/s.

Learn more about speed here: brainly.com/question/6504879

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A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

3 years ago

A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

$F=mg$

$\dfrac{mv^2}{r}=mg$

$\dfrac{v^2}{r}=g$

$v=\sqrt{rg}$

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

$v=\sqrt{250\times9.8}$

$v=49.49\ m/s$

Hence, The speed of space station floor is 49.49 m/s.

6 0

3 years ago

This problem is based on the whole idea of pressure but I’m having trouble on when the area circle formula is included.

Mice21 [21]

Answer:

6.23x10^6Pa

Explanation:

Data obtained from the question include:

F (force) = 490N

r (radius) = 0.005m

A (area of the circlular heel) =?

P (pressure) =.?

First, we'll begin by calculating the area of the circlular heel. This is illustrated below:

Area of circle = πr^2

Area = 22/7 x (0.00)^2

Area = 7.86x10^-5m^2

Pressure is simply force per unit area. It represented mathematically as

Pressure = Force /Area

Pressure = 490/7.86x10^-5

Pressure = 6.23x10^6N/m2

Recall: 1N/m2 = 1Pa

Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa

Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor

8 0

4 years ago

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Doss [256]

Answer: Me non espanol

Explanation:

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3 years ago

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SOVA2 [1]

Answer:

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Explanation:

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6 0

4 years ago