2 years ago

The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter

Physics

2 answers:

-Dominant- [34]2 years ago

4 0

The force of gravity acting on an object is the objects C. weight

ryzh [129]2 years ago

3 0

Answer:

Explanation:

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A web page designer creates an animation in which a dot on a computer screen has a position of r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i

VMariaS [17]

Answer:

V = (5.8cm/s)i, (4.7cm/s)j

Explanation:

Given :

r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^

To obtain the average velocity (V)

V = (r2 - r1) / (t2 - t1)

To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above

r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j

r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j

r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j

r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j

r2 = (16.1cm)i + (9.4cm)j

V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0

V = 11.6i / 2 ; 9.4j / 2

V = (5.8cm/s)i, (4.7cm/s)j

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3 years ago

What conditions must be met in order for work to be done?

Mashcka [7]

Answer:

i think the answer is D

Explanation:

srry if im wrong

4 0

2 years ago

Explain three ways you can get home safely, without getting behind the wheel, if there are drugs or alcohol in your system.

Ad libitum [116K]

Call your parent, call a local taxi ,or a sober friend

4 0

3 years ago

Find equivalent resistance between A and B

V125BC [204]

Answer:

In the picture

Explanation:

I hope that it's a clear solution and explanation, hope that helps.

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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