2. 3. 4. 5. 6. 7. A(n)results when a vehicle loses part or all of its grip on the road. is a technique that can be applied when
1 answer:
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Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
Mobility
We can find the drift velocity replacing,
The electric field is given by,
Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
Answer:
1000 N
Explanation:
First, we need to find the deceleration of the running back, which is given by:
where
v = 0 is his final velocity
u = 5 m/s is his initial velocity
t = 0.5 s is the time taken
Substituting, we have
And now we can calculate the force exerted on the running back, by using Newton's second law:
so, the magnitude of the force is 1000 N.