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3 years ago

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A 1-lb block and a 100-lb block are placed side by side at the top of a frictionless hill. Each is given a very light tap to beg

in their race to the bottom of the hill. In the absence of air resistance, what can you conclude about the outcome of the race?a) The 1-lb block wins the race.
b) The 100-lb block wins the race.
c) The two blocks end in a tie.
d) There’s not enough information to determine which block wins the race

1 answer:

qwelly [4]3 years ago

4 0

Answer:

(c). The two blocks end in a tie

Explanation:

the reason being the absence of any resistance offered to both of the blocks.

if the slope of the hill is for instance 60 deg.

then the acceleration in absence of any resistance is a= 9.81sin(60)

since the acceleration is same then both of the blocks will reach at the same instant

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

2 years ago

PLEASE HELP ASAP!!!<br> GIVING BRAINLIEST!!!<br> 40 points!!

Nana76 [90]

W = Fd = 4(2100) = 8400 J

So the answer is A) 8400 J

I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)

Hope it helps!! :)

4 0

3 years ago

Read 2 more answers

Contrary to popular belief, a ski jumper does not achieve a large amount of "air" when doing a jump (less than 6 feet). This is

Elina [12.6K]

Answer:

The vertical distance that the ski jumper fell is 417.45 m.

Explanation:

Given;

initial horizontal velocity of the jumper, $V_x$ = 26 m/s

horizontal distance of the jumper, dx = 240 m

The time of the motion is given by;

dx = Vₓt

t = dx / Vₓ

t = 240 / 26

t = 9.23 s

The vertical distance traveled by the diver is given by;

$d_y = V_yt + \frac{1}{2}gt^2$

initial vertical velocity, $V_y$ , = 0

$d_y = \frac{1}{2}gt^2\\\\d_y = \frac{1}{2}(9.8)(9.23)^2\\\\d_y = 417.45 \ m$

Therefore, the vertical distance that the ski jumper fell is 417.45 m.

6 0

3 years ago

Which oftthe following is the least reliable source of background information for a scientific project? General internet site, g

kenny6666 [7]

General internet site

7 0

3 years ago

The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate

butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * $10^{8}km$

speed of light = 3* $10^{8}m/s$

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 * $10^{8}*1000$ m

distance between earth and sun = 1.5 * $10^{11}$ m

speed = distance /time

time = distance / speed

time = $\frac{1.5*10^{11} }{3*10^{8} }$

$= 0.5*10^{3}$

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0

3 years ago