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3 years ago

A hobby rocket reaches a height of 72.3 meters and lands 111 meters from the launch point

1 answer:

8 0

Using the following formulas for projectile motion:

Height, H = ( Vo^2 * sin theta^2 )/g

Range, R = ( Vo^2 * sin 2*theta )/g

Rearranging in terms of Vo^2:

Vo^2 = gH / sin theta^2

Vo^2 = gR / sin 2*theta

Equating the two formulas to each other to solve for the angle theta:

gR / sin 2*theta = gH / sin theta^2

Substituting the given values:

(9.8)(111) / sin 2*theta = (9.8)(72.3) / sin theta^2
angle = 52.36 degrees

Therefore, the angle of launch is approximately 52.36 degrees.

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if the efficiency of an electric furnace is 96%, then 96% of the input energy is transformed into thermal energy. what is the ot

Nonamiya [84]

It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.

5 0

3 years ago

Who did not have experimental evidence to support his theory of the atom? Dalton Thomson Rutherford Democritus

Aloiza [94]

Democritus was the one who did not have experimental evidence to support his theory of the atom.

Answer: Option 4

Explanation:

The discovery of atoms were first stated by Democritus but due to the absence of any experimental proof, his statement was not noted as significant at that time.

After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.

This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.

3 0

3 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

4 years ago

Which is an example of reflection

Elena-2011 [213]

Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Mirrors exhibit specular reflection.

5 0

3 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

3 years ago