<u>Answer:</u>
The given statement is a True statement
<u>Explanation:</u>
All cars and trucks have load capacities marked on the jamb of the driver's door, as well as the owner's manual.
This is very significant for braking distance.
A heavier object will require much more braking force and distance to stop, due to New ton's law of motion. “ An object in motion will tend to stay in motion “ .A heavy object will remain in motion longer and require much more force to stop. Four wheel disc brakes are great compared to front disc and rear drum configuration, in dissipating heat, and stopping more efficiently.
Because the specific metals aren’t mentioned in this inquiry.
The educational guesses that we can propose is that:
<span><span>1. </span>The
hypothetical inquiry: There are existing metals for making pots that will cook
food much faster.</span>
<span><span>2. </span>The
one-tailed alternative hypothesis: There are other metals for making pots that
will cook food much faster than the other metals.</span>
<span><span>
3. </span>The
one-tailed null hypothesis: All metals that are used in making pots will cook
food at an equal rate.</span>
To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
