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3 years ago

8

Joe leans against a lamppost with a force of 200 N. The lamppost does not move. How much work does he do?

1 answer:

hoa [83]3 years ago

6 0

Answer:

No work

Explanation:

He is not moving the lamppost as the lamppost cannot move to begin with.

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What happened as the mass of the sun increased during formation of the solor system

sveta [45]

Answer:

Formation. Our solar system formed about 4.5 billion years ago from a dense cloud of interstellar gas and dust. ... When this dust cloud collapsed, it formed a solar nebula—a spinning, swirling disk of material. At the center, gravity pulled more and more material in.

Explanation:

3 0

3 years ago

Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta

e-lub [12.9K]

Answer:

$F_{net} = 1.875\,N$

Explanation:

Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:

$F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]$

$F_{net} = 1.875\,N$

6 0

3 years ago

Mấy bạn việt nam giúp mình với. cần gấp quá

Basile [38]

Saying english so we can help u

6 0

3 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

3 years ago

intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr

V125BC [204]

Answer:

v= u+at

0=45×1000×60×60+3a

_3a=16200045

a=5400015m per sec

4 0

3 years ago

Read 2 more answers