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Elina [12.6K]
3 years ago
12

What will decrease over time as a result of improved

Physics
2 answers:
Semmy [17]3 years ago
5 0

Answer: b

Explanation:

bagirrra123 [75]3 years ago
4 0

Answer:

C

Explanation:

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What system carries food to cells though the blood ?
irinina [24]
"Circulatory system" does that, as all about transportation of blood within the body comes under it's work

Hope this helps!
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3 years ago
Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a
Crank
A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.
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3 years ago
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All magnetic fields result from the movement of
konstantin123 [22]
Don’t know sorry I’m just trying not a good person
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3 years ago
Starting with a constant velocity of 45 km/h, a car accelerates for 35 seconds at an acceleration of 0.45 m/s2 . What is the vel
DENIUS [597]

Answer:

28.3 m/s

Explanation:

Vi = 45 Km/h = 12.5 m/s

Vf - Vi = at

Vf -12.5 = 0.45(35)

Vf= 28.3 m/s

5 0
3 years ago
A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0
Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\

If we carry out summation of Torques on the pulley we will have to,

F_2*d-F_1*d = I \alpha

Where,

I = Inertia moment

\alpha =Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

I = \frac{1}{2} mr^2

Replacing this equations we have know that

(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})

F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}

F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1

Or

F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}

Replacing our values we have that

F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}

F_1 = 0.974 N

Therefore the tension in the string between the pulley and the cart is  0.974 N

6 0
4 years ago
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