The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
1.
m = mass of Mr. Ure = 65 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Ure is given as
F = mg
F = 65 x 9.8
F = 637 N
2.
F = force of gravity on car = 3050 N
m = mass of the car = ?
g = acceleration due to gravity = 9.8 m/s²
force of gravity on car is given as
F = mg
3050 = m (9.8)
m = 3050/9.8
m = 311.22 kg
3.
m = mass of Mr. Rees = 90 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Rees is given as
F = mg
F = 90 x 9.8
F = 882 N
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.
There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:
(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2
We can reduce this to:
(15 x 10)m/36 s^2 = 150 m/36 s^2
Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
Answer:
Temperature after ignition=7883.205 K
Explanation:
The number of moles is,
n=PV/RT
=(1.18x10^6)(47.9x10^-6)/8.314(325)
= 0.0209 moles
a) In this process volume is constant
Q=U
=nCv.dT
dT= Q/nCv
=1970/(1.5x8.314)(0.0209)
= 7558.205 K
The final temperature is,
= 7558.205+325
= 7883.205 K
As I found out the choices for your question which are:
<span>A) F2 to F-
B) Cr2O7²- → Cr2+
C) O2 to H2O
D) HAsO2 to As
</span>
Unfortunately, the answer does not belong to the choices provided. In fact, it is the oxidation half-reaction that occurs at the anode of an electrode for it to transform chemical energy to consumable electrical energy.
