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3 years ago

Help with 3 please! I will give brainliest!

Physics

1 answer:

rusak2 [61]3 years ago

8 0

The answer is point b because vertical velocity is zero at the maximum height

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A rod 14.0 cm long is uniformly charged and has a total charge of 222.0 mc. determine (a) the magnitude and (b) the direction of

Yuri [45]

Electric field due to a charged rod along its axis is given by

$E = \frac{kQ}{(L+r)(r)}$

here we know that

L = 14 cm

r = distance from end of rod

r = 36 - 7 = 29 cm

Q = 222 mC

now we will have

$E = \frac{(9 \times 10^9)(222 \times 10^{-3})}{(0.29)(0.43)}$

$E = 1.6 \times 10^{10} N/C$

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3 years ago

One light-hour is the distance that light travels in an hour. How far is this, in kilometers? (Recall that the speed of light is

ycow [4]

Answer:

B 1.08 BILLION

Explanation:

SEE ATTACHMENT

Download pdf

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2 years ago

Biological factor related to antisocial personality disorder is

dusya [7]

Schizotypal personality disorder and borderline personality disorder

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2 years ago

Read 2 more answers

A 72 kg skydiver is descending on a parachute. His speed is still increasing at 1.2 m/s2. What are the magnitude and direction o

denis23 [38]

Answer:

Parachute exerts a force of 619.2 N upward

The net force is 86.4 N acting downward

Explanation:

As the gravitational acceleration g = 9.8 m/s2, the parachute help reduces the net acceleration to 1.2m/s. So it must exerted an upward acceleration on the skydiver of

9.8 - 1.2 = 8.6 m/s2

Since the skydiver mass is 72 kg, we can use Newton's 2nd law to calculate the force that causes this acceleration of 8.6

F = ma = 8.6*72 = 619.2 N acting upward

The net force is also the product of net acceleration and mass

= 1.2 * 72 = 86.4 N acting downward

5 0

3 years ago

A pilot heads his jet due east. The jet has a speed of 425 mi/h relative to the air. The wind is blowing due north with a speed

Alja [10]

Answer:

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = $\sqrt{425^{2}+35^{2}}$ = 426.88 mi/h

Direction of jet is, ∅ = $tan^{-1} \frac{40}{425}$ = 5.38°

Explanation:

We can represent east direction by i and north direction by j.

The jet has a relative speed of 425 mi/h relative to the air.

The wind is blowing due north with a speed of 35 mi/h = 35 j

425 mi/h is the relative speed with respect to wind that is

Velocity of jet wrt wind= $V_{jw}$ = $V_{j}-V_{w}$

425 i = $V_{j}$ - 35 j

$V_{j}$ = 425 i + 35 j

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = $\sqrt{425^{2}+35^{2}}$ = 426.88 mi/h

Direction of jet is,

∅ = $tan^{-1} \frac{40}{425}$ = 5.38°

7 0

3 years ago