A 60 N force is applied over a distance of 15 m. How much work was done?

ASAP if you answer right I will mark brainiest....

rjkz [21]

Answer: *The inner planet that has two moons is Mars * the only planet in the Solar System with clockwise rotation is Venus * The surface of Mercury is covered with ancient magma ( which is similar to the spews of volcanoes on Earth as well) * The planet that is closest to Earth is Mercury * Venus has more volcanoes than any other planet * Earth's moon formed when a(n) can sometimes be called the Big Splash or can be called the Theia impact (Luna the moon formed from the ejecta of the collision between the Proto- Earth and Mars sized planet) * Mars is called the red planet because it's soil contains the element iron oxide ( which is the compound that gives blood and rust hue)

Explanation: Hope this helped :)

8 0

2 years ago

The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec

PilotLPTM [1.2K]

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

5 0

2 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

7 0

3 years ago

Read 2 more answers

A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a

Fiesta28 [93]

Answer:

I do agree with the student. Because imagene you have a tissue and a stick. And you throw them at the same speed , the stick will go farther because it' s heavier but the tissue would just blow around and not go forward because it is not heavy enought.

Explanation:

Hope this is correct have a terrific day.

5 0

3 years ago