Answer:
<em>Maximum=70 m</em>
<em>Minimum=26 m</em>
Explanation:
<u>Vector Addition
</u>
Since vectors have magnitude and direction, adding them takes into consideration not only the magnitudes but also their respective directions. Two vectors can be totally collaborative, i.e., point to the same direction, or be totally opposite. In the first case, the magnitude of the sum is at maximum. Otherwise, it's at a minimum.
Thus, the maximum magnitude of the sum is 48+22 = 70 m and the minimum magnitude of the sum is 48-22= 26 m
Answer:
Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity = 
Initial vertical velocity= 
(Since ball was thrown horizontally only)
Acceleration acting horizontally, 
= 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration, 
= 9.8 m/s² (Since only gravity acts on it)
Let 't' be the time taken to reach ground
Therefore, using equations of motion, we have
Then using Equations of motion for horizontal motion,
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy
Explanation: i took the test and got a 100%
Answer:
F = 1.047 10⁻² N
Explanation:
Let's use kinematics to find the angular acceleration
w = w₀ + α t
as for rest w₀ = 0
w = α t
α = w / t
let's reduce the magnitudes to the SI system
w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s
m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg
r = 10.0 cm (1 m / 100 cm) = 0.100 m
let's calculate
α = 104.72 / 1
α = 104.72 rad / s²
angular and linear variables are related
a = α r
a = 104.72 0.100
a = 10.47 m / s²
finally we substitute in Newton's second law
F = 1 10⁻³ 10.47
F = 1.047 10⁻² N
