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kati45 [8]
3 years ago
7

A toy rocket is shot straight up into the air with Anitial velocity of 45 m/s

Physics
1 answer:
tangare [24]3 years ago
3 0
Ill save you all the math steps, but here is the answer! <span>102.25m I took that physics exam 3 days ago! So if you need the steps just ask Ill insert them in!</span>
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How does the law of conservation of energy apply to machines?
allsm [11]
Energy can be changed from one form to another, but it cannot be created or destroyed. ... This principle is referred to as the first law of thermodynamics or the law of energy conservation. The law applies to all systems both large and small, and, again, it states that energy cannot be created or destroyed.
4 0
3 years ago
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A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height
krok68 [10]

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

Force = Mass*Acceleration

Acceleration will be gravitational acceleration

Work = (837.6)*(32.17)*(1)

Work = 26945.6 ft⋅lbf

8 0
3 years ago
Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C
DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0
3 years ago
Please help ASAP. In what direction do you need to apply force to move an object vertically?
PilotLPTM [1.2K]
The answer is up . tylrhscjwizn
5 0
3 years ago
During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.

4 0
3 years ago
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