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3 years ago

10

a bowling ball hits two standing bowling pins at the same time. which of the following is true. assume that all collisions are e

lastic

1 answer:

neonofarm [45]3 years ago

6 0

Answer:

b?

Explanation:

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A student on her way to school walks eastward in a straight line 20.0 meters towards the bus stop, but realizes she dropped her

larisa [96]

Answer:

Total displacement will be 47 meter

Total distance will be 83 meters

Explanation:

We have given that first the student go eastward towards bus stop 20 meters

But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters

So displacement = 20-18 = 2 meters

And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters

Total distance traveled by the student = 20+18+45 = 83 meters

3 0

3 years ago

Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.

lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have .21 ( is that down? I will assume it is)

Counterclockwise moments :

.21 * 40 + 1.0 * 20

Clockwise moments :

.5 * 20 + F * 45

these moments must equal each other

.21*40 + 1 *20 = .5 * 20 + F * 45

F = .409 N

7 0

2 years ago

A star that is moving toward an observer has its visible light shifted toward which end of the spectrum? A. blue B. red C. yello

xxTIMURxx [149]

La D mijo es blanco por que al pasar con rapides el color se torna blanco

3 0

3 years ago

Read 2 more answers

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

3 years ago

Read 2 more answers

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago