Answer:

Explanation:
Given:
- angle of launch of projectile from horizontal,

- range of projectile,

<u>We have formula for the range of projectile:</u>

putting the respective values

is the velocity with which Tom should jump to land on the other roof.
Answer:


Explanation:
<u>Given Data:</u>
Weight = W = 65 N
Height = h = 2 m
Time = t = 4 secs
<u>Required:</u>
Power = P = ?
Work Done in the form of Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = Wh
P = P.E. / t
<u>Solution:</u>
P.E. = (65)(2)
P.E = 130 Joules
P = P.E. / t
P = 130 / 4
P = 32.5 Watts
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Hope this helped!
<h3>~AH1807 </h3>
<span>The proper </span><span>battery cable connection when jumping two automotive batteries is : </span><span>(a) negative to negative / positive to positive.
</span><span>Connect the red (positive) cable from the car with the bad battery to the red (positive) on the good battery. </span>
<span>Then connect the black (negative) from the good battery to a grounding point on the other car which should be tightened and metal should be clean.
</span>
<span>Once the car with bad battery has started, the removal of the cable should be in the opposite order. The Red (positive) which was the the First Cable to go on should be the last cable to be taken off.</span>
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.