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2 years ago

How are velocity and acceleration related

Physics

1 answer:

VashaNatasha [74]2 years ago

8 0

Answer:

Acceleration is the rate of change of velocity. If an object is changing its velocity, its speed, or changing its direction, then it is said to be accelerating.

Explanation:

when you change directions or speed up you are accelerating (or decelerating) and when you change directions or speed its called velocity.

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ITS TIMED PLEASE HELP !!!!!

dusya [7]

Answer: direct linear

Explanation:

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3 years ago

Light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb

masya89 [10]

Answer:

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

<em>Hence, the resistance of this bulb is 121 ohms</em>

5 0

3 years ago

Assapp!!!’!!!!!!!! !!!!!!

dlinn [17]

Answer:

delta r(x) = (delta (r)) * cos(alpha), delta r(y) = (delta(r)) * sin(alpha)

Explanation:

Well it's a simple rule I guess...

6 0

3 years ago

An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s? What is the force applied to the object?

valentinak56 [21]

Answer:

10N

Explanation:

Equation: ΣF = ma

Fapp = ma

Fapp = (2kg)(5m/s^2) (im guessing you mean 5.00 m/s^2 not m/s)

Fapp = 10*kg*m/s^2

Fapp = 10N

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2 years ago

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

$Q = ms \Delta T$

$Q = 1*s*(250 - 20)$

$Q = 1*s*230$

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

$Q = ms'\Delta T$

$1*s*230 = 1* s' * \Delta T$

now we have

$\Delta T = (\frac{s}{s'})*230$

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

3 0

3 years ago

Read 2 more answers