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Alexxandr [17]
3 years ago
14

A brick is dropped from a high scaffold. a. How far does the brick fall during this time?

Physics
1 answer:
Alecsey [184]3 years ago
6 0

Answer:

a: after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

after 3 seconds it will have fallen 2.2072

after 4 seconds it will have fallen 3.924

Explanation:

the formula for acceleration due to gravity is (ignoring friction I think)

g = G*M/R^2

earths gravitational constant is about 9.807

g = 9.807*M/R^2

The average weight of a brick is 5 pounds and I'm going to say it's 10 feet off the ground.

g = 9.807*5/10^2.         g = 0.4905 so every second the brick will go 0.4905 fps faster. (fps means feet per second.)

after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

after 3 seconds it will have fallen 2.2072

after 4 seconds it will have fallen 3.924

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Mercury has a mass of 3.3 e 23 kg and a radius of 2.44 e 6 m. Find Mercury's
valentina_108 [34]

Answer:

11.) g = 3.695 m/s^2

12.) g = 8.879 m/s^2

13.) E = 8127 N/C

Explanation:

11.) Given that the

Mercury mass M = 3.3 × 10^23kg

Radius r = 2.44 ×10^6 m

Gravitational constant G = 6.67408 × 10^-11 m3kg-1 s^-2

Gravitational field strength g can be calculated by using the formula below

g = GM/r^2

Substitutes all the parameters into the formula

g = (6.67408 × 10^-11 × 3.3 × 10^23)/(2.44×10^6)^2

g = 2.2×10^13/5.954×10^12

g = 3.695 m/s^2

12.) Given that the

Venus mass M = 4.87×10^24kg

Radius r = 6.05 × 10^6 m

Using the same formula for gravitational field strength g

g = GM/R2

Substitute all the parameters into the formula

g = (6.67408 × 10^-11 × 4.87×10^24)/(6.05×10^6)^2

g = 3.25×10^14/3.66×10^13

g = 8.879 m/s^2

13.) Given that the

Charge = 2.26 nC = 2.26×10^-9

Distance d = 0.05m

Electric field strength E can be calculated by using the formula below

E = Kq/d^2

Where

K = electrostatic constant 8.99 × 10^9 Nm2/C2

Substitutes all the parameters into the formula

E = (8.99 × 10^9 × 2.26×10^-9)/0.05^2

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7 0
4 years ago
Wrist joint is known gliding joint * 1 point true or False ​
zepelin [54]

True

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4 0
3 years ago
Read 2 more answers
Tony drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Tony drove hom
polet [3.4K]

Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

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Distance =y×(a+b)

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ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

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Distance= a×x=ax

Now, applying the questions

So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

a=4×22/(6-4)

a=88/2

a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

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Answer:

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Explanation:

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