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aliina [53]
3 years ago
11

Is this a testable question?

Physics
2 answers:
faltersainse [42]3 years ago
8 0
This is a non testable question because it cannot be answered by doing an experiment. But it could be modified for example Dogs are more obedient then cats.
Stells [14]3 years ago
4 0
Not really; it's primarily a matter of opinion. 
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Explain what effect losing a gamma ray has on a nucleus.
gogolik [260]

Answer:

Hi, Since energy levels in the nucleus are much higher than those in the gas, the nucleus will cool down by emitting a more energetic electromagnetic wave called a gamma ray . Gamma ray emission causes no change in the number of particles in the nucleus meaning both the atomic number and mass number remain the same.

5 0
2 years ago
The periodic table lists all the elements according to the number of __________ they have! You can’t change __________ without c
raketka [301]

Answer:

protons

atomic number

Note: The same atomic number can be associated with several different values of atomic mass, but an element can have only 1 atomic number,

5 0
2 years ago
Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t
Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

4 0
2 years ago
A sprinter with a mass of 80 kg accelerates from 0 m/s to 9 m/s in 3 s. What is the runner's acceleration?
sertanlavr [38]
Acceleration = (change in speed) / (time for the change) = 9/3 = <em>3 m/s²</em> .

His mass makes no difference.
5 0
3 years ago
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
3 years ago
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