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Sedaia [141]
3 years ago
9

If you are good at activities that require agility, what are you able to do well?

Physics
1 answer:
Mrac [35]3 years ago
5 0

Answer:

2

Explanation:

agility is the ability to zoom fast and do fast zoomy stuff

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Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
Juli2301 [7.4K]

Answer:

<h2>23.33 kg </h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m  = \frac{42}{1.8}   = 23.3333... \\

We have the final answer as

<h3>23.33 kg</h3>

Hope this helps you

4 0
2 years ago
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A sound wave travels through water at 1,500 m/s. How far would it travel in 5 s?
hichkok12 [17]
7500m because 1,500 in 1 sec so 1500x5 7500
8 0
3 years ago
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The mass of a basketball is three times greater than a softball. Compare the momentum’s of a softball and a basketball if they b
ella [17]
<span>The momentum of the basketball is three times that of the softball. Momentum equals mass times velocity. Therefore, if the basketball and softball are moving at the same velocity, and the basketball has three times the mass of the softball, the basketball has three times the momentum of the softball.</span>
3 0
3 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi
Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ *  6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

7 0
4 years ago
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