Answer:
Explanation:
The given function is
Now h = the height from the surface of the Earth
Here the building is 458 m tall
So,
Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the
1 answer:
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Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Missing part of the question: determine the magnitude of Porsha's acceleration.
Given the data in the question;
- Mass of Porsha;
- Mass of Zorn;
- Force of Porsha push;
Magnitude of Porsha's acceleration;
To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:
Where m is the mass of the object and a is the acceleration.
We substitute the mass of Porsha and the force he used into the equation
Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Learn more: brainly.com/question/25125444
Answer:
Option (c).
Explanation:
An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,
is the angle of projection
The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.
As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same .
Answer:
i. The radius 'r' of the electron's path is 4.23 × m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r =
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 × ÷ 2.608 ×
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f =
= (1.6 × × 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 × ÷ 5.7263 ×
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.