Answer:
Explanation:
As we know that the formula of range is given as
now we know that
maximum value of the range of the projectile is given as
now we need to find such angles for which the range is half the maximum value
so we will have
A metal block has a density of 5000 kg per cubic meter and a mass of 15,000 kg. What is its volume?
1 answer:
Taking into account the definition of density, the volume of the metal block is 3 m³.
<h3>What is density</h3>Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.
In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.
Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.
<h3>Volume of the metal block</h3>In this case, you know that:
- Density= 5000
- Mass= 15000 kg
- Volume= ?
Replacing in the definition of density:
Solving:
volume×5000 = 15000 kg
volume=
<u><em>volume= 3 m³</em></u>
In summary, the volume of the metal block is 3 m³.
Learn more about density:
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Answer:
B. +5.75 m/s
Explanation:
When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are and
respectively, the relative velocity of B with respect to A is given by:
Thus, the velocity of the girl relative to the lawnmower is:
Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
