Answer:
<h2>2.5 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula

f is the force
m is the mass
From the question we have

We have the final answer as
<h3>2.5 m/s²</h3>
Hope this helps you
Answer: Given that:
Mass (m) = 20 g = 0.02 Kg,
temperature (T₁) = -10°C = -10+273 = 263 K
temperature (T₂) = 10°C = 10+273 = 283 K
Specific heat of water (Cp) = 4.187 KJ/Kg k
We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)
= 0.02 × 4.187 × (283-263)
Q = 1.67 KJ
Heat transferred is 1.67 KJ
Explanation:
Hi there!
a)
When the motor is first turned on, the coils are initially stationary. Thus, there is no change in magnetic flux and, consequently, no induced emf.
Therefore:

Since there's no back emf:

Solving for i using Ohm's Law:

b)
We are given that at max speed, the back EMF is 70.0 V.
Using the same equation as above:

Plugging in the values:

Solving for current:
