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3 years ago

What is the momentum of a ten ton truck which is sitting still?

Physics

1 answer:

ANTONII [103]3 years ago

7 0

The momentum of the truck is given by:

p = mv

p = momentum, m = mass, v = velocity

There isn't too much work to do here; the truck is at rest, therefore its velocity is 0. This means the product of m and v is 0, giving the momentum as 0.

You might be interested in

Which type of reaction does this diagram represent?

Sever21 [200]

Nuclear fusion because atomic nuclei combine to form a heavier nucleus. Option A is correct.

<h3>What is nuclear fusion?</h3>

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.

The more energy it takes to liberate an electron from a smaller atom. This is referred to as binding energy.

As a result, when two little nuclei fuse together, there is more binding energy than when two big nuclei fuse together.

For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.

Nuclear fusion because atomic nuclei combine to form a heavier nucleus.

Hence, option A is correct.

To learn more about nuclear fusion refer to the link;

brainly.com/question/14019172

#SPJ1

8 0

2 years ago

An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (

vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg

angular speed of the disk = 4 rev/s

arm outstretched = 1 m

centripetal force of the disk in the circular path

F = m ω² r

ω = 4 x 2 x π

ω = 25.13 rad/s

F = m ω² r

F = 2 x 25.13² x 1

F = 1263.03 N

hence, centripetal force equal to the F = 1263.03 N

6 0

3 years ago

Electric resistance ____________ with a(n) ____________in the length of a wire and as a result current flow ___________.

maria [59]

The answer is D. Electric resistance increases with an increase in the length of a wire and as a result current flow decreases. There is a direct relationship between the length of the wire and the resistance. The longer the wire, the more resistance there will be. Additionally, from Ohm's Law, current is inversely proportional to resistance. This means as the current increases, resistance decreases or vice versa.

3 0

3 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

3 years ago