2 years ago

So lost on this one number 1 ?

Physics

1 answer:

kirza4 [7]2 years ago

5 0

Read each statement, one at a time, and form the picture in your mind.

<em>Statement #1</em> should make you feel all warm and fuzzy inside.

Statements #2, #3, and #4 should make your red flag wave on the inside, and should make you giggle on the outside when you realize how absurd each of those is.

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With what minimum speed must you toss a 110 g ball straight up to just touch the 11-m-high roof of the gymnasium if you release

Sholpan [36]

Answer:

v = 13.79 m/s

Explanation:

given,

mass of ball = 110 g

height = 11 m

ball is released from = 1.3 m

minimum speed = ?

using conservation of energy

Potential energy is conserved in the form of kinetic energy

$P E =KE$

$m g h= \dfrac{1}{2}mv^2$

$v^2 = 2 g h$

$v=\sqrt{2 g h}$

$v=\sqrt{2\times 9.8 \times (11-1.3)}$

$v=\sqrt{2\times 9.8 \times 9.7}$

$v=\sqrt{190.12}$

v = 13.79 m/s

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3 years ago

What is a discussion?

kap26 [50]

Answer:

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Explanation:

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3 years ago

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The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

monitta

Answer:

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Explanation:

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2 years ago

What did the scientists deduce from the fact that the ants eyes of the desert have multiple lenses

Evgesh-ka [11]

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2 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

3 0

2 years ago