<span>A. </span>Let’s
say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the
components are v0x and v0y.
<span>v0y = 0 since the customer slides it horizontally so applied
force is in the x component only.
<span>The equations for horizontal and vertical projectile motion
are:
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>
Setting the origin to be the end corner of the
counter so that x0=0 and y0=0, hence:
x = v0x t
y = - 1/2 g t^2
Given value are: x=1.50m and y=-1.15m (y is
negative since mug is going down)
<span>1.50m = v0x t
----> v0x= 1.50/t</span>
<span>-1.15m = -(1/2) (9.81) t^2 -----> t =0.4842 s</span>
Calculating for v0x:
v0x = 3.10 m/s
<span>B. </span>v0x
is constant since there are no other horizontal forces so, v0x=vx=3.10m/s
vy can be calculated from the formula:
<span>vy = v0y + at where a=-g
(negative since going down)</span>
vy = -gt = -9.81 (0.4842)
vy = -4.75 m/s
Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy
tan(90-Ø )= 3.1/4.75
Ø =
56.87˚<span> below the horizontal</span>
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
<span>Density can be calculated and found by dividing the sample's mass by its volume. D=m/v</span>
the football player has speed
Answer:
this is were you get everything
Explanation:
