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GarryVolchara [31]
3 years ago
14

Most high mountain ranges are composed of

Physics
1 answer:
mixer [17]3 years ago
6 0

Answer:

folded sediments , undeformed granite, flat-lying sediments, glacial till and metamorphic rock

Explanation:

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2. A ladder 16cm long weighs 420N and it's centre of gravity is 7m from one end.It is carried horizontally by two boys,each hold
ipn [44]

Answer:

well the answers are 180 and 240 obviously

Explanation:

Download pdf
8 0
2 years ago
Read 2 more answers
A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?
Lyrx [107]
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.

To calculate acceleration, divide your weight by the mass, thus the accleration is:

100N/100kg = 1(m/s^2)
7 0
3 years ago
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Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
3 years ago
Using the same cost and time estimates, consider the time-effectiveness of each engineer.
sammy [17]

Answer:

Camilla

Explanation:

I got it right on edge. :)

4 0
3 years ago
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86
Vladimir [108]

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

3 0
4 years ago
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