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12345 [234]
3 years ago
14

I need help with the following three physics problems please!

Physics
1 answer:
omeli [17]3 years ago
3 0
None of these is a Physics problem. The numbers, and the fractions that
they appear in, might have come from Physics problems, but these are
nothing more than arithmetic exercises, and they can be whipped out
in a hurry with a little bit of fancy calculator work.  You don't have to
know any Physics in order to answer any of these. 

The first one is 2.7 x 10⁻²⁴ Newton.
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A rocket is dropped out of an airplane at 100 m/s (downward). If the rocket fires causing an upward acceleration of Ct2 and it t
Mkey [24]

Answer:

C = 0.0125 m/s⁴. The calculation procedure can be found in the attachment below. The concept of motion along a straight line with constant acceleration has been applied to solve the problem.

Explanation:

The sign convention chosen in this problem solution is upwards as positive and downwards negative. The equation of motion v = u + at has been used to calculate the constant C as only one unknown is contained in this equation. This is so because we have been given the initial velocity, the acceleration and the time taken. To solve future problems of this kind, first thing to check for is an equation of motion with the least number of unknown. This helps to reduce the complexity of the problem solution.  

5 0
3 years ago
When you jump upwards, why don't you notice the Earth moving<br> downwards?
mixer [17]

Becuase Gravity. The Earthp pulls you down, you do not pull the earth.

6 0
3 years ago
What is the potential difference (voltage) if the resistance of 20 ohms produces a current of 10 amperes?
JulijaS [17]

Answer:

200 volts

Explanation:

Ohm's Law: V = IR

I = 10 A

R = 20 ohms

V = (10)(20) = 200 volts

8 0
2 years ago
2) A constant net force acts on an object. Describe the motion of the object.?
77julia77 [94]
Constant acceleration
7 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

s=720.3\ m

We need to calculate the velocity

Using formula of velocity

v=a\times t

Put the value into the formula

v=29.4\times7

v=205.8\ m/s

We need to calculate the height

Using formula of height

H=\dfrac{v^2}{2g}

Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

Using formula for maximum height

H'=H+s

Put the value into the formula

H'=2160.9+720.3

H'=2881.2\ m

Hence, The maximum height is 2881.2 m.

4 0
3 years ago
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