Pentru aceleasi corp,pe pamant sau pe luna,difera masa greutatea corpului?

A 50 kg child runs off a dock at 2.0 ms (horizontally) and lands in a waiting rowboat of mass 150 kg. At what speed does the row

Alex73 [517]

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

5 0

4 years ago

The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?

Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

$mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}$

Put all the values,

$h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m$

So, it will reach to a height of 2.5 m.

8 0

3 years ago

The distance between earth and sun is 15000000km. Light takes 499 seconds to reach earth from sun. Calculate the speed of light

iVinArrow [24]

To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

$\rm{Speed = \dfrac{Distance}{Time}$

The speed of the light is 30,060.12 km/sec.

Given to us

The distance between the earth and the sun is 15000000km
Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

$\rm{Speed = \dfrac{Distance}{Time}$

Therefore,

<h3>What is the speed of the light?</h3>

$\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}$

Substitute the value,

$\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}$

$\text{Speed of light} = 30,060.12\ km/sec$

Hence, the speed of the light is 30,060.12 km/sec.

Learn more about Speed, distance, and Time:

brainly.com/question/15100898

7 0

2 years ago

Read 2 more answers

Newton’s laws do not apply to small objects?

Soloha48 [4]

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

7 0

3 years ago

A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is

Veronika [31]

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

$F_c = W-N$

$m\omega^2r = W-N$

Here,

m = Net mass

$\omega$ = Angular velocity

r = Radius

W = Weight

N = Normal Force

$m\omega^2r = 174-146$

The net mass is equivalent to

$F = mg \rightarrow m = \frac{F}{g}$

Then,

$m = \frac{174lb}{32.17ft/s^2}$

Replacing we have then,

$(\frac{174lb}{32.17ft/s^2})\omega^2 (54ft) =174lb-146lb$

Solving to find the angular velocity we have,

$\omega = 0.309rad/s$

Therefore the angular velocity is 0.309rad/s

6 0

3 years ago