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3 years ago

Help in this question, please.

Physics

1 answer:

Mice21 [21]3 years ago

4 0

C. Is the brightest ... It's connected directly across the battery. A. and B. are equal and not as bright as C ... They're in series across the battery. D. doesn't light at all ... It has a wire connected between its ends, so no voltage across it.

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A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

vazorg [7]

Answer:

Explanation:

It is 8z

3 0

4 years ago

Winds blowing toward the east are called easterlies. True or False

eimsori [14]

Answer: False

Explanation:

Winds are named for the cardinal direction they blow from. Hence, a wind that "blows towards the east", logically should come from the west and is called a "west wind".

In thise sense, one of the best examples of this type of wind are the Westerlies, which are are prevailing winds that blow from the west at midlatitudes and have the characteristic that are stronger during winter and weaker during summer.

Therefore, the statement is false.

6 0

3 years ago

If you know this please help

Ainat [17]

Answer:

Explanation:

It's b I sis thebksqjwnsx0qkqnsnd991isnd

3 0

3 years ago

Read 2 more answers

The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)

trapecia [35]

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r ⇒ v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

8 0

3 years ago

A car is moving southwest at a velocity of 10 meters per second. Five seconds later, it's going 40 meters per second. What was t

Inga [223]

If "during this time" refers to the 5 second interval mentioned above, then the average acceleration is

$\bar a=\dfrac{v-v_0}{t-t_0}=\dfrac{40\,\frac{\mathrm m}{\mathrm s}-10\,\frac{\mathrm m}{\mathrm s}}{5\,\mathrm s-0\,\mathrm s}$

Notice that we took the start time to be the start of the 5 second interval and set that to $t_0=0$ . The starting velocity $v_0$ is the velocity measured at the start of the interval, and $v$ is the velocity measured at its end.

So the average velocity over these 5 seconds is

$\bar a=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}{5\,\mathrm s}=6\,\dfrac{\mathrm m}{\mathrm s^2}$

6 0

4 years ago