1 year ago

Can anyone help me on the summer and winter Gizmo questions. Activities A and B?

Physics

1 answer:

Anuta_ua [19.1K]1 year ago

6 0

Answer:

sure, do you have more detailed of the assignments...?

Explanation:

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Robyn has been working out five days a week for the past six months. She is now planning to take a month-long break from the gym

Vsevolod [243]

The answer is C -
because reversibility means when taking a break it'll be hard to get back on schedule

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3 years ago

5. A bus starting from rest accelerates in a straight line at a constant rate of 3m/s2 for 8s. Calculate the distance travelled

mojhsa [17]

Answer:

d = 96 meters

Explanation:

a = acceleration

t = time

d = distance

$d = \frac{1}{2} \times a \times {t}^{2}$

$d = \frac{1}{2} \times 3 \times {8}^{2}$

$d = 96$

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2 years ago

One consequence of Newton's third law of motion is that __________. A. Every object that has mass has inertia B. A force acting

vovangra [49]

Answer:

Explanation:

a force acting upon an object increases that objects acceleration

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2 years ago

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

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2 years ago

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the

natulia [17]

Answer:

The value is $A = 2.80 *10^{-4} \ m^2$

Explanation:

From the question we are told that

The operating temperature is $T = 2450 \ K$

The emissivity is $e = 0.350$

The power rating is $P = 200 \ W$

Generally the area is mathematically represented as

$A = \frac{P}{ e * \sigma * T^2}$

Where $\sigma$ is the Stefan Boltzmann constant with value

$\sigma = 5.67 *10^{-8} \ W/m^2\cdot K^4$

$A = \frac{200}{0.350 * 5.67*10^{-8} * 2450^{4}}$

$A = 2.80 *10^{-4} \ m^2$

8 0

2 years ago