An OSU linebacker of mass 110.0 kg sacks a UM quarterback of mass 85.0 kg. Just after they collide, they are momentarily stuck t
1 answer:
Answer:
The initial speed of the linebacker just before the collision was 4.6 m/s.
Explanation:
Mass of OSU linebacker, m = 110 kg
Mass of UM quarterback, m' = 85 kg
Just after they collide, they are momentarily stuck together, the common speed is, V = 2.6 m/s
Initial speed of quarterback is 0 as it was at rest initially, u' = 0
Let u was the initial speed of the linebacker just before the collision. As they struck together, the momentum remains conserved. So,
So, the initial speed of the linebacker just before the collision was 4.6 m/s.
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Answer:
h = 10.4 m
R = 22.48 m
v= 16,2 m/s , α = 61.7°, below the horizontal
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)
Explanation:
The ball describes a parabolic path, and the equations of the movement are:
Equation of the uniform rectilinear motion (horizontal ) :
x = vx*t :
Equations of the uniformly accelerated rectilinear motion of upward (vertical ).
y = (v₀y)*t - (1/2)*g*t² Equation (2)
vfy² = v₀y² -2gy Equation (3)
vfy = v₀y -gt Equation (4)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m
y: vertical position in meters (m)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Known data
y= 8.8 m
v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s
g = 9.8 m/s²
Calculation of the initial vertical velocity ( v₀y)
We apply Equation (3) with the known data
(vfy)² = (v₀y)² -2*g*y
(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)
(5.7)²+ 172.48 = (v₀y)²
v₀y = 14.3 m/s
Calculation of the maximum height the ball rise (h)
In the maximum height vfy=0
We apply the Equation (3) :
(vfy)² = (v₀y)² -2*g*y
0 = (14.3)² - 2*98*h
h = (14.3)² / 19.6
h = 10.4 m
Calculation of the time it takes for the ball to the maximum height
We apply the Equation (4) :
vfy = v₀y -gt
0 = v₀y -gt
gt = v₀y
t = v₀y/g
t = 14.3/9.8
t= 1.46 s
Flight time = 2t = 2.92 s
Total horizontal distance traveled by the ball (R)
We replace data in the equation (1)
x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)
R = (7.7)* (2.92) = 22.48 m
Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground
vx = 7.7 m/s
vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s
v= 16,2 m/s
α = -61.7°
α = 61.7°, below the horizontal
i- j components of the v
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)
Answer:
The values is
The direction is out of the plane
Explanation:
From the question we are told that
The magnitude of the electric field is
The magnitude of the magnetic field is mathematically represented as
where c is the speed of light with value
Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )
Answer:
The translational kinetic energy is 225 J
The rotational kinetic energy is 225 J
Explanation:
Given;
mass of the wheel, m = 2-kg
linear speed of the wheel, v = 15 m/s
Transnational kinetic energy is calculated as;
E = ¹/₂MV²
where;
M is mass of the moving object
V is the velocity of the object
E = ¹/₂ x 2 x (15)²
E = 225 J
Rotational kinetic energy is calculated as;
E = ¹/₂Iω²
where;
I is moment of inertia
ω is angular velocity
E = ¹/₂ x 2 x (15)²
E = 225 J
Thus, the translational kinetic energy is equal to rotational kinetic energy