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morpeh [17]
2 years ago
11

Do Planets with more mass have more gravity than planets with less mass

Physics
1 answer:
STatiana [176]2 years ago
8 0
Here you go it was too long to type

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Gravity is a <br><br> A.pushing force<br> B. not a force at all<br> C. pulling force
melamori03 [73]

Answer:

C

Explanation:

gravity is a pulling force according to Newton

4 0
2 years ago
How many cubic feet are in a 55-gallon drum?
san4es73 [151]
1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches

                       (55 gal) x (231 in³/gal) x (1 ft³/1728 in³)

                   =    (55 x 231 / 1728)  ft³

                   =         7.352 cubic feet       (rounded)  
5 0
3 years ago
The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

7 0
3 years ago
At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec
evablogger [386]

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

4 0
3 years ago
39. Draw a complete free body diagram of a 40 kg plastic crate at rest on a wooden table (us=0.7). The applied force to the righ
Leya [2.2K]

In order to draw the free body diagram, first let's calculate the friction force acting on the crate:

\begin{gathered} F_f=N\cdot\mu \\ F_f=40\cdot9.8\cdot0.7 \\ F_f=274.4\text{ N} \end{gathered}

Since the friction force is greater than the force applied, the crate will not move, and the friction force will be equal to the force applied.

The weight force is equal to 40 * 9.8 = 392 N.

So, drawing the diagram, we have:

4 0
1 year ago
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