Answer:

Explanation:
<u>Given:</u>
- Mass,
- Velocity,

where,
are the uncertainties in mass and velocity respectively.
The kinetic energy is given by

The uncertainty in kinetic energy is given as:

Answer:
A) B = 0.009185 T
B) Drection is negative y-direction
Explanation:
A) We are given;
Speed(v) = 2.5 x 10^(7) m/s
Acceleration (a) = 2.2 x 10^(13) m/s²
We also know that charge of proton(q) = 1.6 x 10^(-19)
Mass of proton(m) = 1.67 x 10^(-27)
Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;
F = qvBsinθ = ma
Since perpendicular, θ = 90°
And so, sinθ = sin 90 = 1
Thus, qvB = ma
Making B the subject gives;
B = ma/qv
B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))
= 0.009185 T
B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction
Answer:the summery is ...i really don't know the picture is blurry and I cant see can you make it clear?
Explanation:
Answer:
A. W = 6875.0 J.
B. W = -14264.6 J.
Explanation:
A. The work done by the rider can be calculated by using the following equation:

Where:
: is the force done by the rider = 25 N
d: is the distance = 275 m
θ: is the angle between the applied force and the distance
Since the applied force is in the same direction of the motion, the angle is zero.

Hence, the rider does a work of 6875.0 J on the bike.
B. The work done by the force of gravity on the bike is the following:
The force of gravity is given by the weight of the bike.
And the angle between the force of gravity and the direction of motion is 180°.
The minus sign is because the force of gravity is in the opposite direction to the motion direction.
Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.
I hope it helps you!
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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