Initial speed of the train = 7 m/s
Final speed of the train = 17 m/s
Change of speed of the train = (17 - 7) m/s
= 10 m/s
Time taken for the change of the speed of the train = 120 s
Then
Acceleration of the train = Change of speed of the train/Time taken for the change of speed
= 10/120 m/s^2
= 1/12 m/s^2
= 0.083 m/s^2
So the acceleration of the train is 1/12 meter per second square or 0.083 m/s^2. I hope this is the answer you were looking for.
Mass of a model car you measured = 230 grams
Actual mass = 218 grams
Percent error = ?
First subtract the actual value from the value
you measured,
230 grams – 218 grams = 12 grams
Now divide 12 grams by the actual mass that is
218 grams:
12 / 218 = 0.055
Now multiply 0.055 with 100 to get the percent
error;
0.055 x 100 = 5.5%
So, the Percent error is 5.5%.
Answer:
The height of the Everest mountain is, x = 8514.087 m
Explanation:
Given data,
The gravitational field strength at the top of mount Everest, gₓ = 9.772 m/s²
The formula for gravitational field strength is,
<em> gₓ = GM/(R+x)²</em>
Where, x is the height from the surface of the Earth
Therefore,
(R+x)² = GM/gₓ
x = √(GM/gₓ) - R
Substituting the values,
x = √(6.67408 x 10⁻¹¹ X 5.972 x 10²⁴ / 9.772) - 6.378 x 10⁶
x = 8514.087 m
Therefore, the height of the Everest mountain is, x = 8514.087 m
Answer:
r₂ = 4 r
Explanation:
For this exercise let's use Newton's second law with the magnetic force
F = q v x B
bold letters indicate vectors, the magnitude of this expression is
F = q v B sin θ
in this case we assume that the angle is 90º between the speed and the magnetic field.
If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal
a = v² / r
let's use Newton's second law
F = ma
q v B = m v² / r
r = 
Let's apply this expression to our case.
Proton 1
r = \frac{qB_1}{mv_1}
Proton 2
r₂ = 
in the exercise indicate some relationships between the two protons
* v₁ = 2 v₂
v₂ = v₁ / 2
* B₂ = 2B₁
we substitute
r₂ = 
r₂ = 4 
r₂ = 4 r
Answer:
Velocity of throwing = 34.335 m/s
Explanation:
Time taken by the tennis ball to reach maximum height, t = 0.5 x 7 = 3.5 seconds.
Let the initial velocity be u, we have acceleration due to gravity, a = -9.81 m/s² and final velocity = 0 m/s
Equation of motion result we have v = u + at
Substituting
0 = u - 9.81 x 3.5
u = 34.335 m/s
Velocity of throwing = 34.335 m/s
