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3 years ago

I NEED HELP ASAP IM ON TIMER !!!!

Physics

1 answer:

Angelina_Jolie [31]3 years ago

8 0

Answer:

a.) 20 km east

b.) 50 km west

c.) 80 km west

Explanation:

As given,

Railway station - A B C

Distance(km) - 0 30 60

Starts from A , train reaches B be in 15 minutes

Starts from A , train reaches C be in 30 minutes

a.)

If A is origin

As train covers 30km = 15 minutes

⇒ 15 minutes = 30 km

1 minute = $\frac{30}{15} = 2$ km

⇒ 10 minutes = 2×10 = 20 km

∴ we get

The position of the train moving from station A to station B after 10 minutes = 20 km east

b.)

If B is origin then , the position of the train moving from station A to station B after 10 minutes = 30 + 20 = 50 km west

( Because B is the origin , so firstly train go from B to A in 15 minutes ( 30 km ) then A to B in 10 minutes (20 km) )

c.)

If C is the origin then , the position of the train moving from station A to station B after 10 minutes = 60 + 20 = 80 km west

( Because C is the origin , so firstly train go from C to A in 30 minutes ( 60 km ) then A to B in 10 minutes (20 km) )

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2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

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7 0

3 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$