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3 years ago

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A garbage truck crashes head-on into a Volkswagen and the two come to rest in a cloud of flies. Which experiences the greater fo

rce?
a.) the truck

b.) Volkswagen

c.) both experience the same force

d.) not enough information

e.) none of these

1 answer:

Artist 52 [7]3 years ago

8 0

The volkswagon experiences a greater force as it was hit by a larger object that had a larger mass than itself

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A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air $y$ at time $t$ is given by

$y=100\,\mathrm m-\dfrac g2t^2$

Let $d$ be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

$-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}$

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3 years ago

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s isJ. (Formula: )

tekilochka [14]

25 Joule
Formula=.5*mass*velocity^2

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3 years ago

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At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec

evablogger [386]

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out= 3.14 inch³/sec

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3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

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3 years ago

You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity

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The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

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3 years ago