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Delicious77 [7]
4 years ago
13

During the removal of wastes in a cell, the waste products move

Physics
1 answer:
marin [14]4 years ago
7 0
<span>From an area of higher conentration to an area of lower concentration which is B. my friend. Hoped i helped.

</span>

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A 4.5-kg object oscillates on a horizontal spring with an amplitude of 3.8 cm. Its maximum acceleration is 26 m/s2 . Find (a) th
zimovet [89]

Answer:

F = - K X     force constant for spring

a = F / m      maximum acceleration

F = 4.5 kg * 26 m/s^2 = 117 Newtons

(A)  K = 117 N / .038 m = 3079 N/m

ω = (K/M)^1/2 = (117/5)^1/2 = 4.84 / sec

(B) f = ω / 2 pi = 4.84 / 6.28 = .77 /sec

(C) P = 1 / f = 1/.77   = 1.30 sec

3 0
2 years ago
I NEED HELP WITH THE LAST QUESTION ASAP D':
VMariaS [17]
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6 0
3 years ago
14) A race car accelerates from rest to a velocity of 41 m/s in 480 m. What is the acceleration of the race car?
natka813 [3]

Answer:

Explanation:

V₀ = 0 m/s

V = 41 m/s

S = 480 m

___________

a - ?

Distance traveled:

S =  (V² - V₀²) / (2*a)

Acceleration :

a = (V² - V₀²) / (2*S)

a = (41² - 0²) / (2*480) = 1.75 m/s²

8 0
1 year ago
A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?
olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a  ---> a= F/m

a = 40 N/20 kg                   <em>  1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

<em>a= 2 m/s2</em>

7 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
3 years ago
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