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Alexandra [31]
3 years ago
9

You are drinking from a standard (10 oz) coffee cup when a fly starts buzzing around your hands. After finishing your beverage,

you quickly trap the fly inside the cup. You then notice another fly is buzzing nearby and you deftly manage to trap the second fly in the cup without releasing the first fly. If you were able to continue trapping flies until the cup was full, estimate how many flies you could trap in the cup.
Physics
1 answer:
geniusboy [140]3 years ago
6 0
I would estimate over about 25 flies
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A sled starts off with an initial velocity of 8 m/s and slows down to 2 m/s after 3 seconds. What was its acceleration?
melomori [17]

Answer:

-2m/s

Explanation:

3 0
2 years ago
A 29 foot ladder leans against a building so that the angle between the ground and the ladder is 75 ∘ . How high does the ladder
Aloiza [94]

Answer:

28.01m

Explanation:

Opp/Hyp = Sin

Sin 75 = x/29

x = 29 sin 75

x = 28.01m

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3 years ago
A seesaw is 4.0m long with a pivot at its midpoint. A boy who weighs 400N sits at a distance of 1.5m from the pivot. His sister
bezimeni [28]

Answer:

the girl must sit 2 cm from the pivot at the opposite end of the seesaw.

Explanation:

Given;

length of the seesaw, L = 4.0 m

weight of the boy, W₁ = 400 N

position of the boy from the pivot, d₁ = 1.5 m

weight of her sister, W₂ = 300 N

First, make a sketch of this information given;

                 0---0.5m---------------------Δ--------------------------4m

                         ↓<--------1.5m-------> <---------x--------->↓

                        400 N                                          300N

Apply the principle of moment about the pivot, to determine the value of x;

Sum of anticlockwise moment = sum of clockwise moment

400(1.5) = 300(x)

600 = 300x

x = 600/300

x = 2 cm

Thus, the girl must sit 2 cm from the pivot at the opposite end of the seesaw.

8 0
3 years ago
What is the frequency of a wave if the speed is 4300 m/s and the wavelength is 32 m?
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8 0
3 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
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