Answer:
ball clears the net
Explanation:
= initial speed of launch of the ball = 20 ms^{-1}
= angle of launch = 5 deg
Consider the motion of the ball along the horizontal direction
= initial velocity = 
= time of travel
= horizontal displacement of the ball to reach the net = 7 m
Since there is no acceleration along the horizontal direction, we have
Eq-1
Consider the motion of the ball along the vertical direction
= initial velocity = 
= time of travel
= Initial position of the ball at the time of launch = 2 m
= Final position of the ball at time "t"
= acceleration in down direction = - 9.8 ms⁻²
Along the vertical direction , position at any time is given as
Since Y > 1 m
hence the ball clears the net
Answer:
The change in gravitational potential energy is 45 J.
Explanation:
Given that,
Mass = 3 kg
Distance = 1.5 m
Gravitational field strength = 10 N/kg
We need to calculate the change in gravitational potential energy
Using formula of gravitational potential energy
Put the value into the formula
Hence, The change in gravitational potential energy is 45 J.
Calculate for the x and y-components of the velocities involved in this item.
2 mi/h (45° east)
x-component = (2 mi/h)(sin 45°)
= 1.41 mi/h
y-component = (2 mil/h)(cos 45°)
= 1.41 mi/h
4 mi/h (east)
x-component = 4 mi/h
y-component = 0 mi/h
Adding up the corresponding components:
x-component = 5.4142 mi/h
y-component = 1.4142 mi/h
Calculating for the resultant,
R = sqrt ((x²) + (y²))
R = sqrt ((5.4142 mi/h)² + (1.4142 mi/h)²)
R = 5.60 mi/h
Answer: 5.6 mi/h
Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa
"At ground level, ozone contributes to smog" so it is also an air pollutant.
Option: A
<u>Explanation</u>:
ozone is naturally present in stratosphere and acts as shield against harmful ultraviolet radiations. But it acts a pollutant contributing to global warming when it is present in lower level atmosphere particularly troposphere. In this level it combines with primary pollutants that is "nitrogen oxides" and "volatile organic" compounds to form secondary pollutant which absorbs outgoing radiation and contributes in raising the temperature. It has harmful impacts on vegetation as well as human health.
