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3 years ago

Scientific fact can never change 1)True 2)False

Physics

2 answers:

AlexFokin [52]3 years ago

8 0

Answer:

True.

Explanation:

A scientific fact is a fact because it has been proven correct, and therefore cannot change.

Ksivusya [100]3 years ago

4 0

Answer: 2)False

Explanation:

The scientific fact may include the statement which is based on direct observation of a scientific phenomena, scientific event or scientific process occurring in the natural world. The scientific fact is the output of what scientists believe and understand. The knowledge of scientific fact can change with the passage of time due to implementation of new methodologies, technologies and innovations.

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A ball is thrown downward from the top of a building with an initial speed of 25 m/s.

Bumek [7]

Answer:

h = 69.6 m

Explanation:

Data:

Vo = 25 m/s
t = 2.0 s
g = 9.8 m/s²
h = ?

Formula:

$\boxed{\bold{h=V_{0}*t+\frac{g*(t)^{2}}{2}}}$

Replace and solve:

$\boxed{\bold{h=25\frac{m}{s}*2.0\ s+\frac{9.8\frac{m}{s^{2}}*(2.0\ s)^{2}}{2}}}$

$\boxed{\bold{h=50\frac{m}{s^{2}}+\frac{9.8\frac{m}{s^{2}}*4\ s^{2}}{2}}}$

$\boxed{\bold{h=50\ m+\frac{39.2\ m}{2}}}$

$\boxed{\bold{h=50\ m+19.6\ m}}$

$\boxed{\boxed{\bold{h=69.6\ m}}}$

The building has a height of <u>69.6 meters.</u>

Greetings.

5 0

3 years ago

How are science and technology related?

aalyn [17]

One way you could think about it is that you used science of electronics to develop the technology of a computer
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4 0

3 years ago

Read 2 more answers

You apply a force to a cart. The force that you apply is directed to the right. Assume that there is friction acting on the cart

amid [387]

Answer:

As we know that friction force always oppose the motion of body.In other words we can say that friction act opposite to the direction of motion of the body.It tries to stop the body is body is in motion.The value of friction force varies and reaches up to a maximum value.If the body is in rest condition then friction force will be static friction and the body is in moving position then the friction force will be kinetic friction.

Here given that we apply a force in the right direction then friction will act in the left direction.

4 0

3 years ago

A grinding wheel is spinning with an initial angular velocity +ω0. When its motor is turned off at t=0, it begins to slow down w

Olegator [25]

Answer:

Explanation:

Given

initial angular velocity is $\omega _0$

when motor is turned of it started decelerating with $-\alpha$ angular acceleration

angle turned before coming to halt

$\omega ^2-\omega _0^2=2(\alpha )\theta$

here final angular velocity is zero

$-\omega _0^2=-2\alpha \theta$

$\theta =\frac{\omega _0^2}{2\alpha }$

No of revolutions will be $N=\frac{\theta }{2\pi }$

$N=\frac{\omega _0^2}{4\pi \alpha }$

3 0

4 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago