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3 years ago

Cuál fue la primera derrota que sufrieron los fascistas

Physics

1 answer:

valentinak56 [21]3 years ago

8 0

its c so componets arr r\very fragle im aying this because it needs more then 20 letters

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At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

3 years ago

A scientist that applies the laws of science to the needs of communities is called _____.

s2008m [1.1K]

Answer:

The experimental scientist

5 0

3 years ago

The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th

Scilla [17]

Length of the strings = 2.33 m

3 0

3 years ago

The following equation is an example of decay? 232/90 TH---4/2 HE +228/88 RA?

DaniilM [7]

Answer:

Alpha decay

Explanation:

Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.
Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2.
Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.

5 0

3 years ago

Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is

anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_{1} \ and\ m_{2}$

Let the velocity of cars A and B are denoted as $v_{1} \ and\ v_{2}$

The momentum before collision is-

$p_{i} =m_{1} v_{1} +m_{2} v_{2}$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_{1} \ and\ v'_{2}$