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3 years ago

7

Plssss help Do you guys have any other more interesting, funny or creative ideas of incline problems than something sliding down

an incline?

1 answer:

VashaNatasha [74]3 years ago

8 0

U can always just do the classic roller coaster going up an incline and create some sort of story from that.

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Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?

snow_tiger [21]

Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2

4 0

3 years ago

Read 2 more answers

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

marshall27 [118]

Answer:

The resultant velocity is $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

$M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

$v_L$ is the velocity of Luxury Liner = 20 knots due west

$M_f$ mass of freighter = 60,000

$v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

$v_f = 0 \ knots$

So the equation would be

$M_L *v_L = (M_L + M_f) v_t$

Substituting values

$40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

$v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

$= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction

$v_L = 0 \ knots$

So the equation would be

$M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

Making $v_t_n$ the subject

$v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

$= 6 \ knots$

The resultant velocity is

$v_t = \sqrt{v_t_w^2 + v_t_n^2}$

$= \sqrt{8^2 +6^2}$

$v_t=10 knots$

8 0

3 years ago

A car travels from 20-meters to 60-meters in 10 seconds. Calculate the car's speed.

slamgirl [31]

Answer:

8 m/s

Explanation:

All you have to do here is add 20 and 60 (giving you 80) and dividing by 10 seconds. 80/10= 8 m/s

3 0

3 years ago

Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b

zmey [24]

Answer:

The maximum amount of work is $W = 1563.289 \ J$

Explanation:

From the question we are told that

The temperature of the environment is $T = 280\ K$

The volume of container A is $V_A = 2 m^3$

Initially the number of moles is $n = 1.2 \ moles$

The volume of container B is $V_B = 3.5 \ m^3$

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

$W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

$P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

$W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of $R = 8.314 \ J/mol$

Substituting values we have that

$W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

$W = 1563.289 \ J$

8 0

3 years ago

LOL SOMEONE DO THIS FOR ME!!!

olganol [36]

Build a filter.
Dig water from the ground

4 0

3 years ago