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Eddi Din [679]
3 years ago
8

Fatima is watching her pet cat, Winter, napping in the sun. Fatima is curious about the heart rate of Winter when she is napping

, so she develops this scientific question: Does a cat's heart rate change while it is napping? She decides to develop a hypothesis to test this scientific question. What could Fatima's hypothesis be?
Physics
2 answers:
Vsevolod [243]3 years ago
7 0

Answer:

The correct answer will be-<em>"If a cat is naps, the heart beat rate of cat changes."</em>

Explanation:

A scientific hypothesis is a predictive statement based on the works of an earlier investigation. The hypothesis is considered as a well-educated guess which explains the scientific question in a limit.

The scientific hypothesis is proposed in such a way that it could be tested easily through the experiments which can disprove or prove the hypothesis and explain the question.

In the given question, the predicted hypothesis will be-<em> "If a cat naps, the heartbeat rate of cat changes."</em> which can be measured through the instruments.

svetoff [14.1K]3 years ago
3 0

Answer:

Explanation:

There are two hypotheses she could test:

A cat's heart rate changes while it is napping.

A cat's heart rate does not change while it is napping.

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Acid rain is the direct result of acid released by burning fossil fuels. T/F
disa [49]

Answer : False.

Explanation : Acid rain is the indirect result of burning fossil fuels. As the incomplete combustion of fossil fuels releases sulfur dioxide and nitrous dioxide in the air. When it rains it gets mixed with water and forms nitric acid and sulfuric acid which causes acid rain. Acid rain is harmful.

7 0
3 years ago
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What is the change in velocity of a 22-kg object that experiences a force of 15 N for
vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

3 0
3 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
3 years ago
The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an
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1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
where k_e=8.99\cdot10^{9} Nm^2C^{-2} is the Coulomb's constant.
Substituting numbers, we get 
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