If different groups of scientist have access to the same data, how can they draw different conclusions?

lidiya [134]

Electronssssssssssssssssssssssssssss

5 0

3 years ago

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

an 269 kg object is moved a distance of 1.9 m by a force if 580 j of work is done on the object what is the object acceleration

IrinaK [193]

Answer:

Explanation:

w=f*d=580*1.5=870J

7 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

4 years ago

Which best describes the forces identified by Newton’s third law of motion?

egoroff_w [7]

<span>equal and acting on different objects</span>

4 0

3 years ago

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