Ask question

Ask question

All categories

3 years ago

15

You lift a 50 N object 2 meters off the ground what work did you do on the object

1 answer:

mixer [17]3 years ago

4 0

Work equals force × displacement (distance between initial point and end point is displacement)
if u follow this it becomes
work = 50 × 2 which is equal to 100

comment if u have more questions

You might be interested in

Which of the following statements must be true if the net force on an object is zero? Choose all that apply. a) The object must

Alika [10]

Answer:a & d

Explanation:

Given

Net force on object is zero i.e. $F_{net}=m\cdot a_{net}=0$

$a_{net}=0$

if net acceleration is zero therefore change in velocity is zero i.e. velocity is constant

It is not necessary that when an object is at rest acceleration on it is zero for example during Simple harmonic motion of spring mass system when mass is at maximum amplitude i.e. momentarily at rest then its acceleration is maximum.

3 0

3 years ago

Help!!!!!!!!!!!!!!!!!plz!!!!!!!!!!!!!!!!!!!!!

Zarrin [17]

Wrap wire around B the nails

4 0

3 years ago

Read 2 more answers

A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from

Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of incidence beam striking the water is $\theta = 49.63^o$

c the angle of refraction beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is

$h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

$tan \theta = \frac{100}{85}$

$\theta = tan^{-1}(\frac{100}{85} )$

$= 49.63^o$

According to Snell's law

$\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water = 1.333

$i$ is the angle of incidence

$\mu_{air}$ is the refractive index of air = 1

r is the angle of refraction

Substituting values accordingly

$1.33 * sin (40.37) = 1 * sin(r)$

Making r the subject of the formula

$r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

$= 59.7^o$

looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°

i.e $90 -59.7 = 30.3$ °

From the diagram we see that the height target above sea level can be obtained by this relation

$tan \theta = \frac{h}{214}\\$

Where h is the is the height

$tan(30.3) = \frac{h}{214}$

$h = 214 * tan (30.3)$

$=125.05m$

4 0

3 years ago

How can understanding velocity help to prevent a mid-air collision

dsp73

Answer:

To maintain enough time to prevent a collision, a system operating in air traffic where aircraft speed does not

fall below 100 km/h (most medium-sized UAVs and GA aircraft) will need to be able to detect obstacles which

subtend an arc-width of as small as 0.125 mra

7 0

3 years ago

A boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water. How long is he airborn

svp [43]

Answer:

45

Explanation:

7 0

3 years ago