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3 years ago

Will the rocket go up or will it stay on the ground ? Explain the answer if you can also please and thank you.

Physics

1 answer:

Goryan [66]3 years ago

5 0

Answer:

the rocket will go up

Explanation:

there is less force pulling down on the rocket.

100N > 75N

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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this

Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

$p = \frac{fd}{t} \\ where \: \\p = power \\ f = force \\ d = distance \\ and \: t = time \\ \\ p = \frac{3800 \times 50}{10} \\ p = \frac{190000}{10} \\ p = 19000w$

7 0

3 years ago

Is there carbon atoms in the compound Ca3N2

katen-ka-za [31]

No, there are not any carbon atoms inside this compound.

The compound is $Ca_3N_2$ . This means there are 3 "Ca" atoms and 2 "N" atoms.

Ca is calcium
N is nitrogen

Thus, none of the elements in this compound are carbon, meaning there are no carbon atoms. Let me know if you need any clarifications, thanks!

~ Padoru

3 0

3 years ago

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Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink? Explain your answer.

Rainbow [258]

The answer to that is the rock will sink

8 0

3 years ago

Read 2 more answers

Convert 1erg into joule by dimensional method

Valentin [98]

Answer:

1 * 10^-7 [J]

Explanation:

To solve this problem we must use dimensional analysis.

1 ergos [erg] is equal to 1 * 10^-7 Joules [J]

$1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]$

4 0

3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago