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2 years ago

Hằng số phổ biến chất khí

Physics

1 answer:

padilas [110]2 years ago

8 0

Answer:

please in english......................................

Explanation:

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A box is pulled with a horizontal force of 500N and moves 5m what is work done

dalvyx [7]

The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

4 0

3 years ago

A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that

FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$t=\sqrt{\dfrac{2s}{g}}$

$t=\sqrt{\dfrac{2\times 32}{9.8}}$

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

$v=\dfrac{d}{t}$

$v=\dfrac{11\ m}{2.55\ s}$

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0

3 years ago

If a spaceship has a momentum of 30,000 kg-m/s to the right and a mass of

m_a_m_a [10]

Answer:

75m/s

Explanation:

...................

8 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

2 years ago

What mRNA sequence would result from the following DNA sequence?

Nezavi [6.7K]

Answer:

UAC CUG AGG AUC

Explanation:

The mRNA sequence from ATG GAC TCC TAG DNA sequence would be UAC CUG AGG AUC.

According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.

Hence:

ATG GAC TCC TAG will pair with

UAC CUG AGG AUC

5 0

3 years ago