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3 years ago

Which statements describe scientific laws but not theories or hypotheses? Check all that apply.

Physics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Answer:

First and last one,

<h2>Is considered as theories</h2>

yawa3891 [41]3 years ago

6 0

Answer:

B. They do not provide explanations for why they are true.

C. They are considered to be proven facts.

Explanation:

edge 2021

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A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m

katovenus [111]

Answer:

217.43298 m/s

Explanation:

$m_1$ = Mass of bullet = 19 g

$m_2$ = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

$\theta$ = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s$

As the momentum is conserved

$m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s$

The speed of the bullet is 217.43298 m/s

5 0

4 years ago

I need to know how to solve this problem for an upcoming exam

Nataly_w [17]

W=f*d = 5000N * 10 stories * each 4m = 200 000 J

7 0

3 years ago

Read the scenario below and answer the question that follows. Susan was in her algebra class preparing to take a test. Her instr

Sati [7]

Answer:

reduced performance due to stereotype threat

Explanation:

3 0

3 years ago

A lead fishing weight of mass 0.20 kg is tied to a fishing line that is 0.50 m long. The weight is then whirled in a vertical ci

solong [7]

Answer:

Following are the solution to this question

Explanation:

please find the complete question in the attached file.

In point a:

The answer is "bottom".

In point b:

Using formula:

$T= mg + \frac{m V^2}{2}$

$\to 100= 0.2 \times 9.81 + \frac{0.2 \times V^2}{0.5}$

$\to 100= 1.962+ \frac{0.2 + V^2}{0.5}\\\\\to 100- 1.962= \frac{0.2 + V^2}{0.5}\\\\\to 98.038= \frac{0.2 + V^2}{0.5}\\\\\to 49.019=0.2+V^2\\\\\to 48.819=V^2\\\\ \to 6.987 \ \frac{m}{s}$

6 0

3 years ago

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0

3 years ago