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melamori03 [73]
3 years ago
12

#2 Distance (d) = _______37.5cm____________

Physics
1 answer:
bixtya [17]3 years ago
8 0

Answer:

Im not sure how to explain it but i need points im sorry

Explanation:

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A particle of mass 4.5 × 10-8 kg and charge +5.4 μC is traveling due east. It enters perpendicularly a magnetic field whose magn
egoroff_w [7]

Answer:

0.00970 s

Explanation:

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

Force due to magnetic field = qvB sin θ

q = charge on the particle = 5.4 μC

v = velocity of the charge

B = magnetic field strength = 2.7 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

where w = angular velocity.

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

mvw = qvB

mw = qB

w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)

w = 3.24 × 10² rad/s

w = 324 rad/s

w = (angular displacement)/time

Time = (angular displacement)/w

Angular displacement = π rads (half of a circle; 2π/2)

Time = (π/324) = 0.00970 s

Hope this Helps!!!

4 0
3 years ago
A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J
8 0
3 years ago
HELPPP
AlekseyPX
Friction is caused by the uneven surfaces of touching objects
3 0
2 years ago
How we will solve this question?<br> 36.45cL=______=μL
QveST [7]

Answer:

0.364

I believe... Good luck!

8 0
3 years ago
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra
ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{l}

\mu=\dfrac{0.44}{2}=0.22\ kg/m

T=4L^2f\mu

T=4\times (1.75)^2\times (27.5)^2 \times 0.22

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0
3 years ago
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