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valentinak56 [21]
3 years ago
6

Which hand is negatively charged? ОА ОВ. Ос. OD

Physics
1 answer:
Strike441 [17]3 years ago
8 0
Answer is OB
Answer is OB
Answer is OB
You might be interested in
What is the Kinetic Energy of a 60 kg person on skis traveling 20 m/s?
sattari [20]

Explanation:

equating the parameters into the formula, it's gonna be

= ½ × 60 × 20²

= ½ × 60 × 400

= ½ × 24000

K.E = 12000J

3 0
3 years ago
Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
kow [346]

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

          # turns = 1 turn (24 cm / 0.5 cm)

         # laps = 48 laps

Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

       # lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

          L₀ = 2π R

          L = # turns L₀

          L = # turns 2π R

          L = 48 2π 30

          L = 9047.79 cm

          L = 90.48 m

6 0
4 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
2 years ago
Read 2 more answers
Carol has several test tubes of different liquids. If Carol removes heat from the liquid substances, which of the following is m
Archy [21]

Answer:

d

Explanation:

6 0
3 years ago
4. How much work is done when you lift a 25 N box up to a 2 m shelf?
Semmy [17]

Answer:

<h2>50 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 25 N

distance = 2 m

We have

workdone = 25 × 2 = 50

We have the final answer as

<h3>50 J</h3>

Hope this helps you

6 0
3 years ago
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