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3 years ago

12

A soccer ball initially at rest rolls down a hill with an acceleration of 3.0 m /s2 .it accelerates for 5 seconds before melissa

is able to catch it how far did it roll ?

1 answer:

slavikrds [6]3 years ago

6 0

The correct answer is the Soccer ball is 3.2 plus 5 that’s your answer.

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I need help with this!! I dont have to textbook that goes with it and i cant figure anything out :(

sasho [114]

i think number 1 on the bottom is friction, i honestly am having trouble myself so i cannot really help

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3 years ago

Bands of waves in the electromagnetic spectrum are arranged in order of

OLEGan [10]

Bands of waves in the electromagnetic spectrum are arranged in order of <span>frequency and wavelength. The answer is letter A. The rest of the choices do not answer the question above.</span>

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4 years ago

What is the period, in seconds, that corresponds to each of

Charra [1.4K]

Answer:

0.1s,5s,0.017s

Explanation:

T=1÷frequency

7 0

3 years ago

Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius R = 7.5 cm. Assume that the powder

Ipatiy [6.2K]

Answer:

A) $E =\frac{\rho r}{2\epsilon}$

B) $v = 58.7923\times 10^4 V$

Explanation:

a) using Guass law

$\oint E.dA = \frac{q_{enclosed}}{\epsilon_o}$

$EA = \frac{q_{enclosed}}{\epsilon_o}$

$E = \frac{q_{enclosed}}{A \epsilon_o}$

Here $A is = 2\pi rL$

$E = \frac{q_{enclosed}}{(2\pi rL) \epsilon_o}$

Volume charge density is given as

$\rho = \frac{q_{enclosed}}{volume}$

net charge is given as

$q_{enclosed} = \rho \times volume$

therefore $E = \frac{ \rho \times (L\pi r^2)}{(2\pi rL) \epsilon_o}$

$VOLUME = L\pi r^2$

After solving electric field equation we get

$E =\frac{\rho r}{2\epsilon}$

b) electric potential difference is given as

$v_{wall} - v = - \int_{r}^{R} Edr$

$0 - v = - \int_{r}^{R} E dr$

$v = \int_{r}^{R} Edr$

$= \int_{r}^{R} (\frac{\rho r}{2\epsilon}) dr$

$= \frac{\rho}{4\epsilon} (R^2 - r^2)$

at r = 0

$v = \frac{- 3.7 \times 10^{-3} \times 0.075^2}{4\times (8.85\times 10^{-12}}$

$v = 58.7923\times 10^4 V$

7 0

3 years ago

A woman of mass 44 kg jumps off the bow of

Aneli [31]

Answer:

v₃ = 1.59 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of linear momentum.

That is, the momentum is conserved before and after the jump.

Before the jump, we have the mass of the woman and the canoe without speed (at rest). After the jump you have the woman moving to the right and the canoe moving to the left.

$(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})$

where:

m₁ = mass of the woman = 44 [kg]

m₂ = mass of the canoe = 69 [kg]

v₁ = velocity of the canoe at rest = 0

v₂ = velocity of the woman after jumping = 2.5 [m/s]

v₃ = velocity of the canoe after jumping = [m/s]

$(44+69)*0= (44*2.5)-(69*v_{3})\\110 = 69*v_{3}\\v_{3}=1.59[m/s]$

4 0

3 years ago