Explanation:
It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m
Answer:
- 20 j or 20 N/m
Explanation:
this is the write answer I am writing with full of confidence which makes easier in your study
It’s red shift and blue shift because This only occurs when the frequency of the wave is made longer or shorter due to the movement of the source relative to the observer
Answer: At 34°c
Explanation:
Using The Arrhenius Equation:
k = Ae − Ea/RT
k represents rate constant
A represents frequency factor and is constant
R represents gas constant which is = 8.31J/K/mol
Ea represents the activation energy
T represents the absolute temperature.
By taking the natural log of both sides,
ln k = ln A- Ea/RT
Reactions at temperatures T1 and T2 can be written as;
ln k1= ln A− Ea/RT1
ln k2= ln A− Ea/RT2
Therefore,
ln(k1/k2) = −Ea/RT1 + Ea/RT2
Since k2=2k1 this becomes:
ln(1/2) = Ea/R*[1/T2 − 1/T1]
Theefore,
-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]
1/T2 - 1/293 = -1.55 x 10^-4
1/T2 = -1.55 x 10^-4 + 34.13x 10^-4
1/T2 = 32.58 x 10^-4
Therefore T2 = 307K
T2 = 307 - 273 = 34 °c
Because there is water everywhere by which the water gets hot and vaporised so that's why near the sea temperature is low.
