Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula
Where;
g = The acceleration due to gravity = 9.81 m/s², we have;
Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
Explanation:
Formula depicting relation between total flux and total charge Q is as follows.
(Gauss's Law)
Putting the given values into the above formula as follows.
Q =
=
=
= -8.4 nC
Therefore, when the unknown charge is q then,
-14.0 nC + 33.0 nC + q = -8.4 nC
q = -27.4 nC
Thus, we can conclude that charge on the third object is -27.4 nC.
To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.
Mathematically it can be defined as
Where
P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
When the number of moles and volume is constant then the expression can be written as
Or in practical terms for this exercise depending on the final temperature:
Our values are given as
Replacing
Therefore the final temperature of the gas is 800K
Answer:
T=1.384×10⁶seconds
Explanation:
Given data
p (Intensity)=1.30 kw/m²
E (Energy)=1.8×10⁹ J
A (Area)=1.00 m²
T (Time required)=?
Solution
E=PT ................eq(i)
where E is energy
P is radiation power
T is time
Radiating Power is given as
P=pA
Where p is intensity
A is Area
Put P=pA in eq(i) we get
E=pAT
T=E/pA