Answer: you can use salt and a bit of heat the
Explanation:put salt on the place you want it to stick then heat it up a littel and it should freeze together
Answer:
80.386 degrees
Explanation:
We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse
The adjacent side = 699ft
The hypotenuse = 1034ft
using cos∅ = Adjacent/hypotenuse
where ∅ is the unknown angle
cos ∅ = 699/1034 = 0.167
∅ = arccos 0.167 = 80.368°
As easy as one can imagine
That's true. The only way to stop an object from radiating energy
is to cool it to absolute zero. Since the temperature of space is
roughly 3 degrees above absolute zero, the atoms or molecules
of every object have some kinetic energy, and the object radiates
some heat.
Of course it also absorbs heat at the same time, mostly from the
huge number of stars shining on it.
Answer: F = 2.1 x 10^-4N
Explanation: Question is incomplete.
The complete question is; A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth’s magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east.
Given parameters; l = 2.5m, I = 1.5A, B = 0.55 guass = 0.55 x 10^-4 Tesla , theta = 90 (from West to East), F = ?
F = BILsin(theta)
F = 0.55 x 10^-4 x 1.5 x 2.5 x sin 90
F = 2.1 x 10^-4 N.
According to right hand rule, it's direction is upward.
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
