The range (maximum horizontal distance) travelled by a projectile is given by the formula
R = V^2(sin 2A)/g
where
R = range
V = initial velocity of ball = 64 ft/sec. (given)
A = angle of launch = 45 degrees (given)
g = acceleration due to gravity = 32.2 ft/sec^2 (constant)
Substituting values,
R = 64^2(sin 2*45)/9.8
R = 127.20 feet
Since the receiver is 60 yards away (180 feet), he will have to travel a distance of 180 - 127.20 = 52.80 feet to catch the ball.
The ball's total travel time is given by the formula
T = 2V(sin A)/g
where all the terms have been previously defined.
Substituting values,
T = 2(64)(sin 45)/32.2
T = 2.82 sec.
Therefore, in order for the receiver to catch the ball, his speed must be equal to
52.80/2.82 = 18.72 ft/sec.
Hope this helps ya
The increase in gravitational potential energy for an object of mass m is given by
where
is the increase in altitude of the object.
In our problem, m=3.0 kg,
and
(approximated value), so we have
Answer:
Explanation:
From the question we are told that
Frequency of light 
Generally the equation for wave is mathematically given as
Where
Generally the equation for a fundamental Frequency is mathematically given as
Therefore the length of the smallest standing wave
Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
It's a homogeneous mixture
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