Answer:
v = 27.3 m/s
Explanation:
Given that
Acceleration ,a= 4.2 m/s²
Time ,t= 6.5 s
Lets take the maximum speed gain by Thomson's= v
We know that ,if acceleration is constant then the speed v is given as
v= u + a t
v=final speed
u=initial speed
a=acceleration
t=time
Here the initial speed of Thomson's ,u = 0 m/s
Now by putting the values in the above equation we get
v= 0 + 4.2 x 6.5 m/s
v = 27.3 m/s
Therefore the maximum speed gain by Thomson will be 27.3 m/s.
Answer:
<em>1</em><em>)</em><em> </em><em>P</em><em>late</em><em> </em><em>tectonic</em><em> </em><em>theory</em>
<em>2</em><em>)</em><em> </em><em>Lithospheric</em>
<em>3</em><em>)</em><em> </em><em>Plates</em>
<em>4</em><em>)</em><em> </em><em>Heat</em>
<em>5</em><em>)</em><em> </em><em>Less</em><em> </em><em>dense</em>
<em>6</em><em>)</em><em> </em><em>Conventional current</em>
<em>7</em><em>)</em><em> </em><em>Magma</em>
<em>8</em><em>)</em><em> </em><em>Ocean</em><em> </em><em>crust</em>
<em>9</em><em>)</em><em> </em><em>Slowly</em>
<em>10</em><em>)</em><em> </em><em>Drifting</em><em> </em><em>away</em><em> </em>
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
A single polarizer will stop 50% of the incoming light.
