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Katarina [22]
4 years ago
6

Acceleration is measured in_________ m g m/s m/s2

Physics
1 answer:
Nana76 [90]4 years ago
4 0
Acceleration is measured in m/s².

Answer: m/s²
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Write the type of energy present in pond water and kerosene?​
vichka [17]
Potential and kinetic energy
3 0
3 years ago
A 50 kg skater pushes off from a wall with a force of 200 N. What is the skater's acceleration?
lutik1710 [3]

Answer:

Explanation:

mass (m) = 50 kg

Force (F) = 200 N

Acceleration (a)= ?

WE know

F = m * a

200 = 50 * a

a = 200 / 50

a = 4 m/s²

Hope it will help :)

3 0
3 years ago
As Merrill watches his finger with both eyes open as he brings his finger closer to his nose, he feels his eye muscles working.
charle [14.2K]

Answer:

Answered

Explanation:

As Merrill watches his finger with both eyes open as he brings his finger closer to his nose, he feels his eye muscles working. This shows that her eyes  Muscles have both accommodation and convergence.

Accommodation and convergence allows us to view objects both near and at far without double vision.  

4 0
3 years ago
Gauss's law: Group of answer choices can always be used to calculate the electric field. relates the electric field throughout s
kvasek [131]

Answer:

relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss Law states that overall electric flux of a closed surface is equivalent right to charge enclosed which is divided by the permittivity. In other words Gauss Law stress that

net electric flux that pass through an hypothetical closed surface is equivalent to overall electric charge present within that closed surface.

The Gauss law can be expressed mathematically as

ϕ = (Q/ϵ0)

Q = total charge within the surface,

ε0 = the electric constant

5 0
3 years ago
The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the elect
RideAnS [48]

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=\dfrac{\lambda}{2\pi \epsilon_o r}

It is clear that the electric field is inversely proportional to the distance. So,

\dfrac{E}{E'}=\dfrac{r'}{r}

E'=\dfrac{Er}{r'}

E'=\dfrac{125\times 3.5}{1.5}  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

5 0
4 years ago
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