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Juli2301 [7.4K]
2 years ago
15

Why is it important to scientists for a hypothesis to be testable? How would the experiment differ if it were not? What is an ex

ample of each type? Write 3-5 sentences
Physics
1 answer:
Verdich [7]2 years ago
3 0

The experiment wouldn't exist if the hypothesis were not testable, because, as it says, the hypothesis is untestable. Thus, no experiment can be done, and the hypothesis will always be neither true nor untrue because of lack of evidence either way.

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Evaluate ( 64800 ms ) 2 to three significant figures and express the answer in Si units. Express your answer using three signifi
Sergeu [11.5K]

Answer:

42.0×10² second²

Explanation:

Here, time is given in milisecond

(64800 ms)²

= 4199040000 ms²

The SI unit is seconds

1 second = 1000 milisecond

1\ milisecond=\frac{1}{1000}\ second

\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2

4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2

42.0×10² second²

6 0
4 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
jolli1 [7]

1) 333.6 C

In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:

E=3.00\cdot 10^6 N/C

The electric field strength at the surface of a charged sphere is given by

E=\frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}

where

\epsilon_0 = 8.85\cdot 10^{-12} C^2/(N^2 m^2) is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

Here we have a cloud of radius

R = 1.00 km = 1000 m

So we can re-arrange the previous equation in order to find the charge on the cloud:

Q=4\pi \epsilon_0 ER^2=4\pi (8.85\cdot 10^{-12})(3.00\cdot 10^6)(1000)^2=333.6 C

2) 2.08\cdot 10^{21} excess electrons

The total charge of the cloud must be (in magnitude)

Q = 333.3 C

We know that one electron carries a charge of

e = 1.6 \cdot 10^{-19}C

The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:

Q=Ne

where

N is the number of excess electrons

Solving for N, we find:

N=\frac{Q}{e}=\frac{333.3 C}{1.6\cdot 10^{-19} C}=2.08\cdot 10^{21}

6 0
3 years ago
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00
den301095 [7]

Answer:

\theta=7^o

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

\vec r_1=\ km

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

\vec r_2=\ km=\ km

Finally, he drives 9.5 km 35° north of east.

\vec r_3=\ km=\ km

The total displacement is

\vec r_t=\ km+\ km+\ km

\vec r_t=\ km

The direction can be calculated with

\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232

\boxed{\theta=7^o}

7 0
3 years ago
Find the pressure exerted by a 3000 N crate that has an area of 2m squared
ipn [44]
Pressure = Force/ Area = 3000/2 = 1500 pascal.
8 0
3 years ago
Which actions most likely cause the domains in a ferromagnetic material to align? Check all that apply. heating the material rub
ryzh [129]

Answer:

i. rubbing the material against a magnet

ii. placing the material in a magnetic field of opposite polarity

iii. placing the material near a strong magnet

Explanation:

Ferromagnetic materials are majorly metals which can be easily attracted by a magnet. ferromagnetic materials are made up of domains, behaves as minute pieces of magnet. They can be rearranged to align when under the influence of an external magnetic field.

The alignment of the domains in a ferromagnetic material can be caused by either of the following: rubbing the material against a magnet, placing the material in a magnetic field of opposite polarity, placing the material near a strong magnet.

8 0
3 years ago
Read 2 more answers
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