Answer:
The laptop battery is capable of supplying 253980 J.
Explanation:
Given;
voltage of the battery, V = 11.4 V
power consumed by the laptop, P = 8.3 W
duration of the battery before depletion, t = 8.5 hours
Determine the amount of energy supplied by the laptop battery within 8.5 hours.
Energy, E = Power x time
Energy, E = 8.3 W x 8.5 h = 70.55 Wh
Energy, E in joules = 70.55 wh x 3600 s/h = 253980 J
Therefore, the amount of energy supplied by the laptop battery within 8.5 hours is 253980 J.
The x and y components of the velocity vector is 17.32 m/s and 10 m/s respectively.
<h3>
What is the x - component of the velocity?</h3>
The x-component of the ball's velocity is the velocity of the ball in the horizontal direction or x-axis.
The velocity of the ball in x-direction is calculated as follows;
Vx = V cosθ
where;
- Vx is the horizontal velocity of the ball
- V is the speed of the ball
- θ is the angle of inclination of the speed
Vx = (20 m/s) x (cos 30)
Vx = 17.32 m/s
The velocity of the ball in y-direction is calculated as follows;
Vy = V sinθ
where;
- Vy is the vertical velocity of the ball
- V is the speed of the ball
- θ is the angle of inclination of the speed
Vy = 20 m/s x sin(30)
Vy = 10 m/s
Learn more about x and y components of velocity here: brainly.com/question/18090230
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Answer ;
<em>Note : Satellite completes one orbit in the same time the earth rotates one about its axis</em>
circumference of the orbit = 2 ×π × 4.23×10⁷ (similar to the earth orbital speed
The time of earth/orbit traveled = 23.93 hrs ,
Speed of the orbit = (distance) ÷ (time)
= circumference ÷ time
= (2 ×π × 4.23×10⁷) ÷ 23.93 (1 hr = 3600 sec.)
= 3085.14 m/s
<em>The orbital velocity is 3085.14 m/s</em>
Answer:
Δt'/ T% = 90.3%
Explanation:
Simple harmonic movement is described by the expression
x = A cos (wt)
we find the time for the two points of motion
x = - 0.3 A
-0.3 A = A cos (w t₁)
w t₁ = cos -1 (-0.3)
remember that angles are in radians
w t₁ = 1.875 rad
x = 0.3 A
0.3 A = A cos w t₂
w t₂ = cos -1 (0.3)
w t₂ = 1,266 rad
Now let's calculate the time of a complete period
x= -A
w t₃ = cos⁻¹ (-1)
w t₃ = π rad
this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period
T = 2 t₃
T = 2π / w s
now we can calculate the fraction of time in the given time interval
Δt / T = (t₁ -t₂) / T
Δt / T = (1,875 - 1,266) / 2pi
Δt / T = 0.0969
This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is
Δt'/ T = 1 - 0.0969
Δt '/ T = 0.903
Δt'/ T% = 90.3%