Answer:
See below
Explanation:
Vertical position is given by
df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)
do =original position = 2 m
vo = original <u>VERTICAL</u> velocity = 0
a = acceleration of gravity = 9.81 m/s^2
THIS BECOMES
0 = 2 + 0 * t - 1/2 ( 9.81)t^2
to show t =<u> .639 seconds to hit the ground </u>
During this .639 seconds it flies horizontally at 10 m/s for a distance of
10 m/s * .639 s =<u> 6.39 m </u>
Answer:
A= 148.92 m/s²
Explanation:
Given that
U(x,y) = (6.00 )x² - (3.75 )y ³
m= 0.04 kg
Now force in the x-direction
Fx= - dU/dx
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dx= 12 x
When x=0.4 m
dU/dx= 12 x 0.4 = 4.8
So we can say that
Fx= - 4.8 N
From Newtons law
F= m a
- 4.8 = 0.04 x a
a = -120 m/s²
Acceleration in x direction ,a = -120 m/s²
In y -direction
F= - dU/dy
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dy = 0 - 3.75 x 3 y²
When y = 0.56 m
dU/dy = - 3.75 x 3 x 0.56 x 0.56
dU/dy = - 3.52
So we can say that force in y -direction
F= 3.52 N
F= m a'
3.52 = 0.04 x a'
a'=88.2 m/s²
acceleration in y direction is 88.2 m/s²
The resultant acceleration
A= 148.92 m/s²
<span>The answer is The conductance of a conductor is inversely
proportional to the cross-sectional area of the conductor.</span>
<span>Conductance is directly related to the ease offered by any material to the passage of electric current. Conductance is the opposite of resistance. The higher the conductance, the lower the resistance and vice versa, the greater the resistance, the less conductance, so both are inversely proportional</span>
Answer:Water Only
Explanation:
Given
vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero
If hot water at 
is mixed with cold water at 
then at equilibrium vessel contains only water and final temperature of water will be between and
Heat released by hot water is equal to heat gain by cold water .
Answer:
Approximately 
, assuming that the acceleration of this ball is constant during the descent.
Explanation:
Assume that the acceleration of this ball, 
, is constant during the entire descent.
Let 
denote the displacement of this ball and let 
denote the duration of the descent. The SUVAT equation 
would apply.
Rearrange this equation to find an expression for the acceleration, , of this ball:
.
Note that 
and 
in this question. Thus:
.
Let 
denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is , the net external force on this ball would be 
.
Since 
and 
, the net external force on this ball would be:
.
