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Lady bird [3.3K]
4 years ago
9

Albert Einstein first hypothesized that a large object, such as a planet, bends the space around it. This hypothesis became part

of the theory of relativity, which is about how objects affect the space and time around them.
What did this hypothesis most likely need before it became part of an accepted theory?
a. other scientists asking the same question
b. other questions about mass and space
c. experimental evidence that mass is not related to space
d. experimental evidence supporting the hypothesis
Physics
2 answers:
Natasha2012 [34]4 years ago
7 0

D) Experimental evidence supporting the hypothesis

pentagon [3]4 years ago
3 0

Answer: The answer is (D)

Explanation:

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On a foggy day, you are driving at 70 mph and Granny is driving at 50 mph. As you try to pass her, you both see an overturned tr
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Answer:

C?

Explanation:

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3 years ago
Please help! This is due in 10 minutes
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8 0
3 years ago
A 0.20 kg mass is oscillating at a small angle from a light string with a period of 0.78 s.
Scorpion4ik [409]

Answer:

L = 15 cm

Explanation:

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(0.78/2π)²

L = 0.151027... m

L = 15 cm

7 0
3 years ago
Read 2 more answers
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is
podryga [215]

Answer: critical angle, sin^-1 (n2/n1)

Explanation: the angle of incidence at which the retracted ray makes an angle of 90° with the normal is known as the critical angle.

Snell's law defined refraction mathematically as shown below

n1 sin θi = n2 sin θr

n1 = refractive index of the first medium

n2 = refractive index of the second medium

θi = angle of incidence

θr = angle of refraction

When the refrafted ray is perpendicular to the normal, the angle of refraction (θr) is 90° hence making the angle of incidence (θi) the critical angle θc

By substituting these conditions into the Snell's law, we have that

n1 sin θc = n2 sin 90

According to trigonometry, the value of sin 90 is 1, hence we have that

n1 sin θc =n2

sin θc = n2/n1

θc = sin^-1 (n2/n1)

3 0
3 years ago
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