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AlladinOne [14]

2 years ago

7

slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu

s

1 answer:

andrew-mc [135]2 years ago

7 0

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

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how much voltage is in the primary coil if there are 3200 winding in the secondary coil 500 winding in the primary and 25v in th

Tema [17]

Answer:

Voltage in the primary winding of the coil is 3.9 V

Explanation:

As we know by the principle of transformer we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 25 volts$

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now we have

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7 0

2 years ago

Rocks A and B are located at the same height on top of a hill. The mass of rock A is twice the mass of rock B. How does the pote

Ronch [10]

Potential energy is defined by formula

$U = mgh$

here

m = mass

g = acceleration due to gravity

h = height

Now here two different stones are located at same height

while mass of stone A is twice that of stone B

so here we can say potential energy of A is

$U_a = (2m)gh$

Similarly potential energy of B is

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now if we take the ratio of two energy

$\frac{U_a}{U_b} = 2$

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7 0

3 years ago

A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

6 0

2 years ago

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morpeh [17]

Answer:

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4 0

2 years ago

Ivanshal [37]

Answer:

The velocity is 60 km/hr.

Explanation:

<h3><u>Given:</u></h3>

Displacement (d) = 480 km = 48000 m

Time (t) = 8 Hours = 480 minute

Velocity (v) = ?

Now,

Velocity = Displacement ÷ Time

v = d/t

v = 480/8

v = 60 km/hr

Thus, The velocity is 60 km/hr.

<u>-TheUnknownScientist 72</u>

5 0

2 years ago