Answer:
1. Distance travelled = 12 km.
2. Displacement = 8.6 km
Explanation:
From the question given above, the following data were obtained:
Distance 1 (d₁) = 7 km
Distance 2 (d₂) = 5 km
Total distance =?
Displacement =?
1. Determination of the distance travelled.
Distance 1 (d₁) = 7 km
Distance 2 (d₂) = 5 km
Total distance (dₜ) =?
dₜ = d₁ + d₂
dₜ = 7 + 5
dₜ = 12 km
2. Determination of the displacement.
In the attached photo, R is the displacement.
We can obtain the value of R by using the pythagoras theory as illustrated below:
R² = 7² + 5²
R² = 49 + 25
R² = 74
Take the square root of both side
R = √74
R = 8.6 km
In a Class One Lever, the Fulcrum is located between the Load and the Force.
Yes, if the temperature increases, than that means the particles are moving faster. Temperature is the measure of movement of particles in an object or substance.
By thermal energy, you mean adding heat correct....? I'm not very good at this sort of thing, but I gave you what I have..
Answer:768.75N
Explanation:
mass=62.5kg
acceleration=12.3m/s^2
Force=mass x acceleration
Force=62.5 x 12.3
Force=768.75N
a) create an expression
for the ball's initial horizontal velocity, V0x, in terms of the variables
given in the problem statement.
v0x = vf * cos(Θf)
<span>
b) calculate the ball's initial vertical velocity, V0y, in
m/s</span>
v0x = 32.4m/s * cos(-25.5º)
= 29.2 m/s <span>
tanΘ = v1y / v0x → tan(-25.5) = v1y / 29.2m/s → v1y = -13.93
m/s
the vertical velocity when the ball was caught.
(v0y)² = (v1y)² + 2as = (-13.93m/s)² + 2 * 9.8m/s² * 5.5m = 301.78
m²/s²
v0y = 17.37 m/s
c) calculate the magnitude of the ball's initial velocity,
v0, in m/s</span>
v0 = sqrt (v0y^2 +
v0x^2)
v0 = sqrt (17.37^2 + 29.2^2)
m/s
v0 = 33.98 m/s
<span>
d) find the angle, theta0, in degrees above the horizontal at
which which the ball left the bat.</span>
tan Θ = v0y/v0x
<span>Θ = arctan(17.37/29.2) =
30.75º above horizontal</span>