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timurjin [86]
2 years ago
9

Explain why a law is accepted as facr, but a theory is not​

Physics
1 answer:
Verizon [17]2 years ago
3 0

Answer:

Explanation:

A law and theory are distinct levels in the scientific method. They do not lead to one another.

A law is a description of an observed phenomenon in the natural world. Laws are always true and do not provide explanations as to why they hold true.

A theory is an explanation of an observed phenomenon. It is usually based on experimental evidence and bounded most time in the limits of available data.

  • A law cannot be deposed. It is a fact and holds true at all times.
  • Theories can be discarded even with new technological advancements that provides a new way of study.
  • This is why it is not a fact.
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slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

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What does the subscript 2 indicate in the compound MgCl2?
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It means there are two chlorine molecules and there is only one magnesium molecule (because no subscript really just means there is an invisible 1 that isn't listed).
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a guitar string is 0.620 m long, and oscillates at 234 Hz. if a player uses his finger to shorten the string to 0.480 m, what is
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