A train travels 120 km in 2 hours and 30 minutes. What is its average speed?
2 answers:
Answer:
<em>The average speed of the train is 48 km/hr</em>
Explanation:
The term average speed refers to how fast a body is covering a distance. it is the rate at which a body covers a distance with time. Cars, Humans, and animals are among bodies that can attain speed. speed is measure in m/s.
<h3>Step by Step Calculation</h3>speed of an object can be obtained using the equation below;
Speed = ....................................1
from the question we are provided with;
Distance = 120 km
Time = 2 hours 30 minutes
Speed =?
The time can not operate with two units, so we have to convert 30 minutes to hours;
60 mins = 1 hr
30 mins = x hrs
x =
x = 0.5 hours
now our time will be 2 hrs + 0.5 hr = 2.5 hrs
Substituting our values in equation 1 we have;
Speed =
Speed = 48 km/hr
<em>Therefore, the average speed of the train is 48 km/hr</em>
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Complete Question:
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.
Answer:
a) Speed of car A at the start of sliding = 4.23 m/s
b) speed of car B at the start of sliding = 3.957 m/s
c) Speed of car B before the collision = 7.28 m/s
Explanation:
NB: The figure is not provided but all the parameters needed to solve the question have been given.
Let the frictional force acting on car A, ............(1)
Since frictional force is a type of force, we are safe to say .......(2)
Equating (1) and (2)
a) Speed of A at the start of the sliding
b) Speed of B at the start of the sliding
Let the speed of car B before collision =
Momentum of car B before collision =
Momentum after collision =
Applying the law of conservation of momentum:
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
2442.5 Nm
Explanation:
Tension, T = 8.57 x 10^2 N
length of rope, l = 8.17 m
y = 0.524 m
h = 2.99 m
According to diagram
Sin θ = (2.99 - 0.524) / 8.17
Sin θ = 0.3018
θ = 17.6°
So, torque about the base of the tree is
Torque = T x Cos θ x 2.99
Torque = 8.57 x 100 x Cos 17.6° x 2.99
Torque = 2442.5 Nm
thus, the torque is 2442.5 Nm.
Answer:
Horizontal component = 16.8 m/s
Vertical component = 46.0 m/s
Explanation:
If we denote the initial velocity by <em>v</em> and the angle above the horizontal by <em>θ</em>,
the horizontal component of this initial velocity is given by
The vertical component is given by