Answer:
a. The horizontal component of acceleration a₁ = 0.68 m/s²
The vertical component of acceleration a₂ = -0.11 m/s²
b. -9.19° = 350.81° from the the positive x-axis
Explanation:
The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ = 17.0i - 1.00j m/s
a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²
The horizontal component of acceleration a₁ = 0.68 m/s²
The vertical component of acceleration a₂ = -0.11 m/s²
b. The direction of the acceleration relative to the unit vector i,
tanθ = a₂/a₁ = -0.11/0.68 = -0.1618
θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis
Answer:3,936.4
Explanation:
If 1 foot is 30.28 centimeters multiply 30.28 by 130
Answer:
<em><u>solution</u></em>
<em>3</em><em>0</em><em>8</em><em>=</em><em>2</em><em>0</em><em> </em><em>swings</em><em> </em>
<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>
<em>(</em><em> </em><em>3</em><em>0</em><em>8</em><em>×</em><em>1</em><em>)</em><em>÷</em><em>2</em><em>0</em>
<em>3</em><em>0</em><em>8</em><em>÷</em><em>2</em><em>0</em>
<em>1</em><em>5</em><em>4</em><em>÷</em><em>1</em><em>0</em>
<em>=</em><em>1</em><em>5</em><em>.</em><em>4</em>
<em>=</em>15.4
Answer:
The gravitational force is 3.509*10^17 times larger than the electrostatic force.
Explanation:
The Newton's law of universal gravitation and Coulombs law are:

Where:
G= 6.674×10^−11 N · (m/kg)2
k = 8.987×10^9 N·m2/C2
We can obtain the ratio of these forces dividing them:
--- (1)
The mass of the moon is 7.347 × 10^22 kilograms
The mass of the earth is 5.972 × 10^24 kg
And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C
Replacing these values in eq1:

Therefore

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field
Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m