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3 years ago

35 miles North 6 feet down 136 MB 85 km SE

Physics

2 answers:

lys-0071 [83]3 years ago

8 0

Answer:

The answer is 136 MB I hope that this helps

Explanation:

pantera1 [17]3 years ago

4 0

The answer is 133 Se

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) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi

natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0

3 years ago

At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a² + b²

r = 101.98 km, a = 20 km, b = 100 km

$2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} ) + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} ) + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 ) + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt} = \frac{1900}{101.98} = 18.63 \ km/h$

5 0

3 years ago

A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during t

irina [24]

Answer:

Weight = 734.46 N

Explanation:

Given:

Initial height of the pole-vaulter is, $h_i=5.31\ m$

Final height of the pole-vaulter is, $h_f=0\ m$

Change in the vaulter's potential energy is, $\Delta U=-3900\ J$

We know that, change in potential energy is given as:

$\Delta U=mg(h_f-h_i)$

Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².

Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,

$\Delta U = w(h_f-h_i)$