All energy,potential kinetic within a specific system

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

3 years ago

two identical-looking, large, round balls are placed in front of you. one is filled with feathers and the other is filled with s

Lelu [443]

If you apply a little bit of force, one will move easier than the other since it is lighter.

3 0

3 years ago

Can someone help me pls

Dmitry_Shevchenko [17]

First question: 800J
Second question: 20.4m

4 0

3 years ago

In Lesson 20, a magnesium strip was used to ignite the thermite reaction. When magnesium is placed in a flame from a small blow

nadezda [96]

Answer:

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

Explanation:

The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.

$[Mg]=1s^22s^22p^63s^2$

In order to attain noble gas configuration it will loose two electrons.

$[Mg]^{2+}=1s^22s^22p^6$

$Mg\rightarrow Mg^{2+}+2e^-$ ...[1]

Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.

$[O]=1s^22s^22p^4$

In order to attain noble gas configuration it will gain two electrons.

$[O]^{2-}=1s^22s^22p^6$

$O+2e^-\rightarrow O^{2-}$ ..[2]

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

6 0

3 years ago

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

3 years ago