A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon
1 answer:
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Answer:
v₀ =3.8 m/s
Explanation:
Newton's second law of the box:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Known data
m=2.1 kg mass of the box
d= 5.4m length of the roof
θ = 20° angle θ of the roof with respect to the horizontal direction
μk= 0.51 : coefficient of kinetic friction between the box and the roof
g = 9.8 m/s² : acceleration due to gravity
Forces acting on the box
We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.
W: Weight of the box : In vertical direction
N : Normal force : perpendicular to the direction the roof
fk : Friction force: parallel to the direction to the roof
Calculated of the weight of the box
W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N
x-y weight components
Wx= Wsin θ= (20.58)*sin(20)° =7.039 N
Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N
Calculated of the Normal force
∑Fy = m*ay ay = 0
N-Wy= 0
N=Wy =19.34 N
Calculated of the Friction force:
fk=μk*N= 0.51* 19.34 N = 9.86 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-f = ( 2.1)*a
7.039 - 9.86 = ( 2.1)*a
-2.821 = ( 2.1)*a
a=(-2.821) /( 2.1)
a= -1.34 m/s²
Kinematics of the box
Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement = 5.4 m
v₀: initial speed
vf: final speed = 0
a : acceleration of the box = -1.34 m/s²
We replace data in the formula (2)
0²=v₀²+2*(-1.34)*(5.4)
2*(1.34)*(5.4)= v₀²
v₀ = 3.8 m/s
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.
where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.
Answer:
7500 m
Explanation:
The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.
We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so
d = 2L
In a time of
Electromagnetic waves travel in a vacuum at the speed of light, which is equal to
Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:
(1)
where
, where L is the distance between the radar and the object
Re-arranging eq(1) and substituting, we find L: