All categories

2 years ago

The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged

Physics

1 answer:

gogolik [260]2 years ago

5 0

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

$W = Fd$

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

$W = (200\ N)(500\ m)\\$

You might be interested in

Which of the following is an example of technology influencing science?

ElenaW [278]

I would say your answer is A.

7 0

2 years ago

Read 2 more answers

The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An

Xelga [282]

Answer:

Explanation:

mass of probe m = 474 Kg

initial speed u = 275 m /s

force acting on it F = 5.6 x 10⁻² N

displacement s = 2.42 x 10⁹ m

A )

initial kinetic energy = 1/2 m u² , m is mass of probe.

= .5 x 474 x 275²

= 17923125 J

B )

work done by engine

= force x displacement

= 5.6 x 10⁻² x 2.42 x 10⁹

= 13.55 x 10⁷ J

C ) Final kinetic energy

= Initial K E + work done by force on it

= 17923125 +13.55 x 10⁷

= 1.79 x 10⁷ + 13.55 x 10⁷

= 15.34 x 10⁷ J

D ) If v be its velocity

1/2 m v² = 15.34 x 10⁷

1/2 x 474 x v² = 15.34 x 10⁷

v² = 64.72 x 10⁴

v = 8.04 x 10² m /s

= 804 m /s

5 0

2 years ago

(01.01 LC) Two boxes are 8 cm apart. Which of the following should Janet do to

Trava [24]

Place the boxes 10cm apart, if they are closer that concentrates the mass of the boxes more

4 0

2 years ago

Read 2 more answers

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

8 0

2 years ago

A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

$r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}$

$r_{cm} = {(37.26, 0)}{14.8}$

$r_{cm} = (2.52 cm, 0)$

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

$r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}$

$r_g_{cm} = {(37.26, 0)}{14.8}$

$r_g_{cm} = (2.52 cm, 0)$

So center of mass is same as center of gravity because value of gravity is constant here

3 0

2 years ago