What kind of small objects composes much of the universe?

A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×

otez555 [7]

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, $m=1.01\times 10^7\ kg$

Force applied on the tracks, $F=7.5\times 10^5\ N$

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

$F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s$

So, the required time is 300.3 seconds.

4 0

3 years ago

SECTION B THEORY QUESTIONS (1a) Define the following (a) Work:

Vesna [10]

Work is the amount of energy transferred

Explanation:

In physics, work is a measure of the energy transfer occurring in a process. Typically, we talk about work when energy is converted from one form into another.

For instance, work is done when a force is applied on an object. The work done on the object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

We notice the following:

No work is done when the force is perpendicular to the displacement ( $cos 90^{\circ}=0$ )
The work is maximum when the force is parallel to the displacement

Whenever work is done, there is also an energy transfer taking place. For instance, in the previous example, when the force is applied to the object, the object will accelerate (assume there is no friction), and will gain kinetic energy: therefore, there is a transfer of energy to the object.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

6 0

3 years ago

A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after

vodka [1.7K]

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

3 0

4 years ago

water flows horizontally from a hose that is 3m above the ground. the water lands 7.2m to the right of the hose . how long does

Sidana [21]

Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.782 s

Round as needed.

8 0

3 years ago

Read 2 more answers

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$