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kobusy [5.1K]
3 years ago
11

Freight trains can produce only relatively small acceleration and decelerations. (a) what is the final velocity (in m/s) of a fr

eight train that accelerates at a rate of 0.0600 m/s2 for 8.00 min, starting with an initial velocity of 5.00 m/s?
Physics
2 answers:
miss Akunina [59]3 years ago
7 0
For this problem, we use the derived equations for rectilinear motion at constant acceleration.

a = (v - v₀)/t
where
a is acceleration
v₀ and v are the initial and final velocities, respectively
x is the distance
t is the time

The solution is as follows:
a = (v - v₀)/t
0.06 m/s² = (v - 5 m/s)/(8 min * 60 s/1 min)
Solving for v,
v = 33.8 m/s

m_a_m_a [10]3 years ago
5 0
(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s 
Acceleration a = 0.0600 m/s² 
time = 8 min = 8 x 60 = 480 s 
Final velocity 
= u + at 
= 5.00 + 0.0600(480) 
= 33.8 m/s

The final velocity is 33.8 m/s
You might be interested in
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0
2 years ago
A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the
zavuch27 [327]
'  +4 m/s² ' means that the pigeon's speed is  4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is 

                           (10 x 4m/s)  =  40 m/s.

We can't say anything about its velocity, because we have
no information regarding the direction of its flight.
5 0
3 years ago
Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

4 0
3 years ago
All pressure topics in physics​
AfilCa [17]
Absolute, Atmospheric, Differential, and Gauge Pressure
5 0
2 years ago
An average man is 80 k (800 N) and the area of the shoes he is wearing is 0.0092 m2. What is the pressure he is exerting on the
S_A_V [24]

Answer:

P = 86956.52 Pa

Explanation:

Data:

  • F = 800 N
  • A = 0.0092 m²
  • P = ?

Use the formula:

  • \boxed{\bold{P=\frac{F}{A}}}

Replace and solve:

  • \boxed{\bold{P=\frac{800\ N}{0.0092\ m^{2}}}}
  • \boxed{\bold{P=86956.52\ Pa}}}

The pressure it exerts on the ground is <u>86956.52 Pascal.</u>

Greetings.

7 0
2 years ago
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