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den301095 [7]
2 years ago
15

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

new acceleration would be _____ m/s/s.
Physics
1 answer:
alexandr402 [8]2 years ago
5 0

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar
finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

\displaystyle V=k\frac{Q}{r}

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

d=\sqrt2 a

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}

\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m

The total potential is  

V_t=V_1+V_2+V_3+V_4

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of \pm 3\mu\ C. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}

V_1=V_2=50912\ V

The total potential is

V_t=50912\ V+50912\ V=1\times 10^5\ V

\boxed{V_t=1\times 10^5\ V}

6 0
2 years ago
A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N
Mnenie [13.5K]

Answer:

8N and 32N

Explanation:

Given that a  light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.

To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.

Let the two upward forces produce by the saw horses be P1 and P2

Assuming that the weight is negligible

Sum of the upward forces = sum of the downward forces.

P1 + P2 = 40 ....... (1)

Also, the sum of the clockwise moment = sum of the anticlockwise moments.

Let's assume that the board is uniform. The weight will act at the centre.

Taking moment at the centre:

P1 × 5 + 40 × 2 = 0

P1 = 40 / 5

P1 = 8N

Substitute P1 into equation 1

8 + P2 = 40

P2 = 40 - 8

P2 = 32N

3 0
2 years ago
Read 2 more answers
A mosquito flying over a highway strikes a windshield of a moving truck. Compared to the magnitude of the force of the truck on
krok68 [10]

Answer:

the answer to this question is

<em>The</em><em> </em><em>Same</em><em> </em>

<em>newton's</em><em> </em><em>law</em><em> </em><em>#3</em>

Explanation:

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em>

3 0
2 years ago
Part A
irina [24]

Answer:

v' = -18 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f} (1)

  • The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

       m_{b} * v_{b} -M_{c}*V_{c}  = m_{b} * 18 m/s + (-M_{c}* 20 m/s)  (2)

  • The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
  • So, we can write again (3) as follows:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (4)

  • Replacing (2) and (4) in (1), we get:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (5)

  • Simplifying, and rearranging, we can solve for v'b, as follows:
  • v'_{b} = -18 m/s (6), which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had  inverted its speed after the collision.
3 0
1 year ago
¿Cual es la fuerza ejercida por el sol sobre la luna ?
zimovet [89]
El sol le da luz a la luna entonces la luna brilla
8 0
2 years ago
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