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3 years ago

4) The mass of Pluto is 1.31 x 1022 kg and its radius is 1.15 x 106 m. What is the acceleration of

Physics

1 answer:

Elina [12.6K]3 years ago

3 0

Answer:

0.661 m/s²

Explanation:

g = MG / r²

g = (1.31×10²² kg) (6.67×10⁻¹¹ m³/kg/s²) / (1.15×10⁶ m)²

g = 6.61×10⁻¹ m/s²

g = 0.661 m/s²

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A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

$v=3.49 m/s$

7 0

3 years ago

Am 1 years old UwU (JK 15)

igomit [66]

Answer:

so am i ._. 15 not 1 ofc lol

Explanation:

7 0

3 years ago

What resistance would produce a current of 200A with a potiential difference of 200V?

viktelen [127]

Aw, I hate physics, is this on Apex?

Resistance can be calculated with the information given in the question.
Equation for Resistance: R = V/I
V (voltage) = 200 Volts
I (current) = 200 Amps

So 200 divided by 200 = freaking 1

Answer: R = 1 (ohms)

Hope this Helps!

5 0

3 years ago

A 15.7-g aluminum block is warmed to 53.2 °c and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 °

ipn [44]

The equilibrium temperature of aluminium and water is 33.2°C

We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K

Now we can calculate the equilibrium temperature

(mc∆T)_aluminium=(mc∆T)_water

15.7*0.9*(53.2-T)=32.5*1*(T-24.5)

T=33.2°C

6 0

3 years ago

Read 2 more answers

The question states: two large, parallel conducting plates are 12cm

AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, $E=\frac{F}{q}$

$E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}$

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0

3 years ago