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yawa3891 [41]
3 years ago
12

I need to know the answer to the question

Physics
1 answer:
nasty-shy [4]3 years ago
7 0
Mostly a hope this helped
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488 J of work is done to a box which is moved across the floor for a distance of 8.9 m. What net force is required to act on the
kvv77 [185]
According to the formula
a = f \times d
Where a is work, f is force and d is the distance that box was moved over. And from that formula, you can get that f = a/d and that is 54.83N of force
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3 years ago
Explain, using numbered steps, how you could find the volume of a small, irregularly shaped rock.
dmitriy555 [2]
<span>1. Get a graduated cylinder. 2. Fill the graduated cylinder to a known amount of water. Record the amount of water in the cylinder. 3. Place rock into the graduated cylinder 4. Measure the new volume of the graduated cylinder with the rock in it. 5. Take the difference of the new volume and the old volume and that is the volume of the rock.</span>
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The velocity of an object is equal to the distance divided by time. The equation is velocity = distance/time. If you wanted to c
Korolek [52]

Answer: C) divide: distance ÷ velocity

Explanation:

The velocity V equation is distance d divided by time t:

V=\frac{d}{t}

If we isolate t we will have:

t=\frac{d}{V}

Hence, the correct option is C: distance divided by velocity.

7 0
3 years ago
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4 0
3 years ago
Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin
Arte-miy333 [17]

Answer:

x=\dfrac{r}{\sqrt2}

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

E=K\dfrac{Q}{(r^2+x^2)^{3/2}}

For maximum condition

\dfrac{dE}{dx}=0

E=K{Q}{(r^2+x^2)^{-3/2}}

\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}

For maximum condition

\dfrac{dE}{dx}=0

K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0

r^2+x^2-3x^2=0

x=\dfrac{r}{\sqrt2}

At x=\dfrac{r}{\sqrt2} the electric field will be maximum.

3 0
3 years ago
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