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e-lub [12.9K]
3 years ago
14

What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between the

m is 440 N
Physics
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

2.87m

Explanation:

Using the law of gravitation to solve this question

F = GMm/r²

G is the gravitational constant

M and m are the masses

r is the distance between the masses

Substitute the given values

G = 6.67×10^-11 m³/kgs²

M =8.8 x 10^6 kg

m = 5.6 x 10^5 kg

F =440N

400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²

400r² = 328.698×10

400r² = 3286.98

r² = 3286.98/400

r² = 8.21745

r = √8.21745

r = 2.87m

Hence the distance of separation is 2.87m

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A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One
Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\

To get the required work done, we will divide the mass by the speed of one knot to have:

w=\frac{7230}{0.51}\\w= 14,176.47Joules

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0
2 years ago
plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu
tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

where

\Delta V is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

E=5.00\cdot 10^6 V/m

d = 2.50 mm = 2.50\cdot 10^{-3} m

Substituting, we find

\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V

3 0
3 years ago
Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.4 m below. The droplets are fal
expeople1 [14]

Answer:

The third drop is 0.26m

Explanation:

The drop 1 impacts at time T is given by:

T=sqrt(2h/g)

T= sqrt[(2×2.4)/9.8]

T= sqrt(4.8/9.8)

T= sqrt(0.4898)

T= 0.70seconds

4th drops starts at dT=0.70/3= 0.23seconds

The interval between the drops is 0.23seconds

Third drop will fall at t= 0.23

h=1/2gt^2

h= 1/2×9.81×(0.23)^2

h= 0.26m

4 0
3 years ago
(100 pts)<br>Why can't you use distance divide by time to calculate the instantaneous<br>speed?​
Fiesta28 [93]

Answer:

Instantaneous speed means speed at any instant

that means Speed is changing with time

You know speed is distance/time

So that means distance is also changing with time

So we take infinitesimal small distance per infinitesimal small time As we assume speed is constant in infinitesimal small time dt

So, we take speed = ds/dt

ds = infinitesimal small distance

dt = infinitesimal small time

As its ratio is equal to speed at any instant

Note : We are taking infinitesimal small distance

But :) we are taking infinitesimal small time also

As you know if denominator is small fraction is large So fraction always give large value

So it's not O ( this makes confuse to most of students)

So, thanks

Good question

Keep thinking like this :)

4 0
3 years ago
Read 2 more answers
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
uysha [10]

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

3 0
3 years ago
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