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Arte-miy333 [17]
4 years ago
9

I have a pure breed homozygous long claw monster (CC), Claude and

Physics
1 answer:
Colt1911 [192]4 years ago
8 0
Idkndkfkfkfrkkffkg. Kdirkrif nejrnrbd idleness r t t y y t r y y y r r d f y y t td
You might be interested in
How does the carbon rod of a cell beomes a positive<br>terminal?​
ivann1987 [24]

Answer:

In the middle of a dry cell, is a rod made of carbon. Around the carbon rod is a chemical paste. At the same time, the carbon rod becomes positively charged. When this happens, electrical current flows out of the cell when a conductor is attached between the cell's positive and negative terminals.

7 0
3 years ago
A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
GalinKa [24]

Answer:

The electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Find the electric flux

Electric flux is calculated using the following formula;

Ф = q/ε

Where ε is the electric constant permitivitty

ε = 8.8542 * 10^{-12}

Substitute ε = 8.8542 * 10^{-12} and q =7.6\µC; The formula becomes

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Hence, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
3 years ago
2. What would be the acceleration of the clown at 5 s? (A) 1.6 m/s2 (B) 8.0 m/s2 (C) 2.0 m/s2 (D) 3.4 m/s2 3. After 12 seconds,
Neko [114]
Is there an image that goes with this question?
7 0
2 years ago
A coin placed on the cover of a book just begins to move when the cover makes an angle of 38 degrees with the horizontal. What i
RSB [31]

Answer:

m g sin theta = force of object along incline due to gravity

N μ = frictional of incline on object where N is the normal force

N = m g cos theta     force perpendicular to incline

m g sin theta = N μ = μ m g cos theta

μ  = tan theta = tan 38 = .78

6 0
2 years ago
A shell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. Negl
Triss [41]

Answer:

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g............................. Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

<em>Given: u = 1700 m/s, </em>∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

8 0
3 years ago
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