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solniwko [45]
2 years ago
9

True or False. Alfred Wegener contributed greatly to the theory of plate tectonics by collecting and analyzing fossil, glacial a

nd geologic evidence.
Physics
2 answers:
bearhunter [10]2 years ago
8 0

Answer:

The Answer would be True

Explanation:

Fed [463]2 years ago
5 0

Answer:

The answer is true

Explanation:

It is true

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Why is it important for scientists to replicate each other’s experiments?
Vilka [71]

Explanation:

to determine if important scientific results are repeatable

6 0
3 years ago
Read 2 more answers
Starting at 48th Street, Dylan rides his bike due east on Meridian Road with the wind at his back. He rides for 20 min at 15 mph
Nezavi [6.7K]

Answer:

55 min

Explanation:

The missing question is: how long does the trip take?

First of all, we need to find the initial distance covered by Dylan. In the first part, he rides for

t_1 = 20 min = \frac{1}{3}h

at a speed of

v = 15 mph

therefore, the distance he covered is

d = v t_1 = (15)(\frac{1}{3})=5 mi

Then Dylan stopped for a time of

t_2 = 5 min = \frac{5}{60}=\frac{1}{12}h

Finally, on the way back, Dylan covered again this distance but travelling at a new speed of

v = 10 mph

So, the time he took is

t_3 = \frac{d}{v}=\frac{5}{10}=\frac{1}{2}h = 30 min

So, the total time of the trip was

t=t_1 + t_2 + t_3 = 20 min + 5 min + 30 min = 55 min

6 0
3 years ago
A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the
Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end  and another test charge at point p

Potential energy is given by  = \frac{k*Q1*Q2 }{R}

where k is electrostatics constant = 9 *10^9

Q1 = first charge , Q2=  test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = \sqrt{8^2 + 6^2} = 10

New Potential energy

U1 = kq1*q2 / 10

substituting  kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

5 0
3 years ago
What is the time required for an object starting
PtichkaEL [24]
(3) 10.1 second Using equation of motion 500 = (0.5)(9.81)t^2. Rearranging, t = sqrt(1000/9.81) = 10.1s
8 0
4 years ago
If a projectile travels in the air for 8 seconds when does a projectile reach its highest point
emmainna [20.7K]

Given

The projectile is in air for a time of t=8 sec

To find

The time it takes to reach the highest point

Explanation

A projectile moves up to the highest point and then again moves down following a parabolic path.

So it will reach the highest point at a time half the time it requires to follow teh parabolic path.

The time taken to reach the highest point is 4 sec

Conclusion

The time taken is 4 sec.

5 0
1 year ago
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