Answer:
The torque in the coil is 4.9 × 10⁻⁵ N.m
Explanation:
T = NIABsinθ
Where;
T is the torque on the coil
N is the number of loops = 9
I is the current = 7.8 A
A is the area of the circular coil = ?
B is the Earth's magnetic field = 5.5 × 10⁻⁵ T
θ is the angle of inclination = 90 - 56 = 34°
Area of the circular coil is calculated as follows;

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°
T = 4.9 × 10⁻⁵ N.m
Therefore, the torque in the coil is 4.9 × 10⁻⁵ N.m
4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
Ok the velocity of an object in free fall is given by the equation :
v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time.
so, we substitute values into this equation. v=35.8-9.8*2.5; v=11.3 m/s
Answer:
1%
Explanation:
Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

The measured value is 2.02 meters and the actual value is 2.00 meters.


First, evaluate the fraction. Subtract 2.00 from 2.02

Next, divide 0.02 by 2.00

Finally, multiply 0.01 and 100.

The percent error is 1%.
Answer:

Explanation:
-The only relevant force is the electrostatic force
-The formula for the electrostatic force is:

E is the electric field and q is the magnitude of the charge.
#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

-Applying Newton's 2nd Law:



#equate the two forces:

#The equations for velocity in uniform acceleration:

#For the proton:

#For the electron:

The mass values of the proton and electron are:

The speed of the ion is therefore calculated as:

Hence, the ion's speed at the negative plate is 