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Eddi Din [679]
2 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]2 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]2 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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what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t
liq [111]
Recall this equation for a device in a direct current circuit:
P = IV
P is the power dissipated by the device, I is the current through the device, and V is the voltage drop of the device.

If we choose to use the ampere as the unit of current and the volt as the unit of voltage, then the product of the current and the voltage will give the power with watts as the unit.
5 0
2 years ago
A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h
Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

6 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
1 year ago
Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.
creativ13 [48]

Answer: a) 456.66 s ; b) 564.3 m

Explanation: The time spend to cover any distance a constant velocity is given by:

v= distance/time so t=distance/v

The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

replacing in the above expression we have

time 1= 627 s

time2 = 297 s

The distance traveled is 564,3 m

8 0
2 years ago
A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
Mekhanik [1.2K]
The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
7 0
1 year ago
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