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3 years ago

A 2.0 kg mass weighs 10 Newtons on planet X. what is the acceleration due to gravity on planet X? Show the work.

Physics

1 answer:

KengaRu [80]3 years ago

8 0

Answer: The gravitational acceleration on planet X is 5 N/kg

On Earth (with the gravitational accelartion g_E) the mass of 2kg will correspond to

$F_E = m\cdot g_E = 2.0 \mbox{kg}\cdot 9.8 \frac{N}{kg} = 19.6 N$

On planet X we are told the same measure is only 10N. Since there is a proportional relationship between g and F, we can calculate g_X:

$\frac{F_E}{g_E}=\frac{F_X}{g_X} \implies\\g_x = \frac{g_E\cdot F_X}{F_E}=\frac{9.8 N/kg \cdot 10 N }{19.6N}=5 \frac{N}{kg}$

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At what depth of lake water is the pressure equal to 201kpa?

Alex787 [66]

Answer:

20m

Explanation:

Pressure = pgh

p = density of water 1000

kg/m^3

g = acceleration due to gravity 9.81 m/s^2

h is the depth of water

Pressure = 201 kPa = 201 x 10^3 Pa

201 x 10^3 = 1000 x 9.81 x h

201 x 10^3 = 9810h

h = 20.49 m

Approximately 20 m

5 0

2 years ago

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

5 0

2 years ago

What happens to the current supplied by the battery when you add an identical bulb in parallel to the original bulb?

Marysya12 [62]

Answer:

b. The current stays the same.

Explanation:

In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .

Now, when an identical bulb is connected in parallel to the original bulb .

Therefore, both the resistance( bulb) are in parallel.

We know, when two resistance are in parallel , current through them is same and voltage is divided between them.

Therefore, in this case current stays same in the original bulb.

Hence, this is the required solution.

6 0

3 years ago

What is (3.25 x 10^5) + (7.5 x 10^4)?

ivanzaharov [21]

Answer:

400000

Explanation:

So first solve one part:

(3.25 * 10^5)

(3.25 * 100,000)

= 325000

Then solve the next part:

(7.5 * 10^4)

(7.5 * 10000)

= 75000

Now lastly, add the two answers:

325000 + 75000 = 400000

Therefore,

(3.25 x 10^5) + (7.5 x 10^4) = 400000

5 0

2 years ago

Read 2 more answers

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igor_vitrenko [27]

Try and calm the child and say that you won't sit on Erin. If you tell the young girl that Erin isn't real at that age, they may have social issues later.

6 0

2 years ago