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3 years ago

Scientists use _____ as chemical symbols.

2 answers:

7 0

The periodic table, which consists of 109 elements, that we know of, it was existing still before we got the systematic method. Jons Jacob Berzelius assigned these elements in place. The symbols always consist of 1 or 2 letters, first one is capital, and the 2nd one, (if its there) is always lowercase.

Short Answer: Periodic Table

Hope i helped.

love history [14]3 years ago

6 0

A single element on the periodic table

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Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal

kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

3 0

3 years ago

The force that opposes motion to moving parts is _____

Colt1911 [192]

The force that opposes motion to moving parts is F<span>riction</span><span>

Hope this helped!
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6 0

4 years ago

Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated

Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>

Gravitation is the force of attraction between any two bodies.
Every body in the universe attracts every other body with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
Mathematically we can expressed it as,

$F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2$

<h3>How to solve the problem?</h3>

Here, we have given with the data's,

$M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m$

Thus, the force of attraction between these two bodies will be,

$F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN$

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0

2 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

Which of the following is a result of a change in pressure?

ioda

B. Exfoliation. Hope I helped you out bro.

7 0

4 years ago