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kakasveta [241]

3 years ago

15

Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

de R/R, es negativa. ¿cuáles son los ángulos posibles de R, medidos a partir del eje x positivo?

1 answer:

Masteriza [31]3 years ago

8 0

If you can write this in English I can help

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A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

lisabon 2012 [21]

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

= zero    - (43 m/s)

=    -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

      = (-43 / 0.28) (m/sec) / sec

      =    153.57... m/s²

      = 1.5... x 10² m/s² .

5 0

3 years ago

Read 2 more answers

If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?

astraxan [27]

Mass of the object is given as

$m = 1417 g = 1.417 kg$

now the speed of object is given as

$v = \frac{d}{t}$

here we know that

$d = 47 m$

$t = 90 s$

now we will have

$v = \frac{47}{90} = 0.52 m/s$

now we will have kinetic energy of the object as

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(1.417)(0.52)^2$

$KE = 0.19 J$

now the power is defined as rate of energy

so here we can find power as

$P = \frac{KE}{t}$

$P = \frac{0.19}{90} = 2.14\times 10^{-3} W$

so above is the power used for the object

3 0

3 years ago

Compared to a plane gravitational pull on the ground to 7 miles in sky is what

Citrus2011 [14]

When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.

4 0

3 years ago

A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver

MissTica

Answer:

Approximately $2.31\; \rm m \cdot s^{-2}$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Assuming that $g = 9.81\; \rm m \cdot s^{-2}$ the weight on this 72-kg skydiver would be $W = m \cdot g = 72 \; \rm kg \times 9.81\; \rm m \cdot s^{-2} = 706.32\; \rm N$ (points downwards.)

Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.

Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:

$\begin{aligned}F(\text{net force}) &= W - F(\text{air resistance})\\ &= 706.32\; \rm N - 540\; \rm N =166.32\; \rm N \end{aligned}$ .

Apply Newton's Second Law of motion to find the acceleration of this skydiver:

$\begin{aligned}a &= \frac{F(\text{net force})}{m} \\ &= \frac{166.32\; \rm N}{72\; \rm kg} = 2.31\; \rm m \cdot s^{-2} \end{aligned}$ .

5 0

3 years ago

Homogous chromosomes separate during.....?

stepladder [879]

Not entirely sure if you're saying Homologous , but assuming you do , the homologous chromosomes seperate in the anaphase stage of Mitsosis of the Cell cycle

3 0

3 years ago