If the rocket launches with a force of 10N, what is the force on the launch pad?

Physics

1 answer:

ser-zykov [4K]3 years ago

5 0

Answer:

10N

Explanation:

Action and Reaction forces state that the force will be equal

You might be interested in

Is gas matter? How do you know? Hint: Remember there are 2 things we need to be considered matter.

astra-53 [7]

Answer:

Gas is a state of matter that has no fixed shape and no fixed volume.

In addition to solids and liquids, gases are also a physical state in which matter can occur. All gases have weight. Unlike solids and liquids, gases will occupy the entire container that encloses them.

matter is "anything that has mass and volume (occupies space)

Gases have mass. The space between gas particles is empty. Gases can be formed as products in chemical reactions. Gas particles can form bonds between them under certain conditions

Gases have volume which isn't fixed (no fixed volume) and no fixed shape. Gases expand to fill the space available. They can also be compressed into a very small space.

Explanation:

7 0

2 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$