Which vector is the sum of the vectors shown below?

A 60 kg person sits in a chair. How much does the earth pull on the person? (Acceleration due to gravity is -10m/s2) F = mxa → F

MA_775_DIABLO [31]

Answer:

600N(downwards)

Explanations

600N(downwards)

Mas of the person = 60kg

Acceleration due to gravity = -10m/s²

To get the earths pull on the person, we will use the Newton second law of motion;

Force = mass * acceleration;

Force = 60 * -10

Force- -600N

Hence the earth gravitational pull on the person is 600N(downwards). It is downwards due to the negative sign.

7 0

2 years ago

Which of the following terms is best described as the number of waves that pass a point in one second? A) wave speed B) period C

ehidna [41]

The term that best describes how many waves that pass? It's frequency because how many waves are passed by a given point or time is called the waves frequency. I hope this helped you out on your assignment.

3 0

2 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun

Sphinxa [80]

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

a)

- The intensity at the surface of Venus is calculated as:

I_v = P_s / 4*pi*r^2_v

I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

I_v = 2,700 W / m^2

b)

- The intensity at the surface of Mars is calculated as:

I_m = P_s / 4*pi*r^2_m

I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

I_m = 610 W / m^2

c)

- The intensity at the surface of Saturn is calculated as:

I_s = P_s / 4*pi*r^2_s

I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

I_s = 16 W / m^2

7 0

3 years ago

Please help me! It states constant speed but moving in circular motion. I know there is acceleration due to a rate of change of

Morgarella [4.7K]

Direction of resultant force in circular motion is directed towards the center of the circle. Hence vector B is the direction of the resultant force.

8 0

2 years ago