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3 years ago

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A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the accel

eration of the truck as its trailer leaks sand at a constant rate through a hole in its bottom?

1 answer:

pickupchik [31]3 years ago

4 0

Answer:

Acceleration will increase.

Explanation:

The relation between force, mass and acceleration according to the Newton's second law of motion is given as:

F = ma

We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:

$a=\frac{F}{m}$

Since, F is constant, the acceleration of the truck is inversely proportional to the mass.

There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.

Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.

So, the acceleration of the truck will increase.

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A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.

Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\$

assume the bullet goes to right side and the gravitational acceleration =10 $ms^{-2}$

so now the weight of the rifle= $\frac{25}{10}$

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}$

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0

3 years ago

vodomira [7]

Answer:

what are u asking there isnt a question

3 0

3 years ago

A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago

An infrared wave traveling through a vacuum has a frequency of 4.2

ololo11 [35]

Answer:O 7.1x10^-7 m

Explanation:

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3 years ago

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A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie

ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, $A=7.4\ cm^2=7.4\times 10^{-4}\ m^2$

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, $B_1=0.5\ T$

Final magnetic field, $B_2=3\ T$

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_2-B_1}{t}$

$\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}$

$\epsilon=1.65\times 10^{-3}\ V$

Induced current in the loop of wire is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{1.65\times 10^{-3}}{2.4}$

$I=6.87\times 10^{-4}\ A$

So, the induced current in the loop of wire over this time is $6.87\times 10^{-4}\ A$ . Hence, this is the required solution.

7 0

3 years ago