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harkovskaia [24]

3 years ago

13

A moving walkway has a speed of .4 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of 3

.0 m/s to the east. What's the mans velocity relative to the moving walkway?

1 answer:

SVEN [57.7K]3 years ago

8 0

Explanation:

Vnet = 0.4m/s - 3.0m/s = - 2.6m/s

so the man walks 2.6m/s towards the west

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The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what di

ss7ja [257]

Answer:

Explanation:

Here image distance is fixed .

In the first case if v be image distance

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

In the second case

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )

3 0

3 years ago

11. Examine the table provided. Which of the objects is producing

ivann1987 [24]

Answer:

b is the answer tennis ball

3 0

2 years ago

A red train travelling at 72 km/h and a green train travelling at 144 km/h are headed toward each

abruzzese [7]

Answer:

Collision will occur.

Speed of red train when they collide = 0 m/s.

Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

Equation of motion v = u + at

u = 20 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 20 - 1 x t

t = 20 s.

Equation of motion s = ut + 0.5at²

u = 20 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 20 x 20 - 0.5 x 1 x 20² = 200 m

So red train travel 200 m before coming to stop.

For green train:-

Equation of motion v = u + at

u = 40 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 40 - 1 x t

t = 40 s.

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 40 s

a = -1 m/s²

Substituting

s = 40 x 40 - 0.5 x 1 x 40² = 800 m

So green train travel 800 m before coming to stop.

Total distance traveled = 800 + 200 = 1000 m>950 m.

So both trains collide.

Distance traveled by green train when red train stops(t=20s)

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 40 x 20 - 0.5 x 1 x 20² = 600 m

Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

Speed of red train when they collide = 0 m/s.

Distance traveled by green train when they collide = 950 - 200 = 750 m

Equation of motion v² = u² + 2as

u = 40 m/s

s= 750 m

a = -1 m/s²

Substituting

v² = 40² - 2 x 1 x 750 = 100

v = 10 m/s

Speed of green train when they collide = 10 m/s.

6 0

3 years ago

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

= 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

= $0.5 x 25 x 5.6^{2}$ = 392 J

(C) average frictional force = $\frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}$

change in KE (ΔKE) = initial KE - final KE
ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$
when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

\frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

= $\frac{(-392 + 2940)}{12}$ = 212.33 N

5 0

3 years ago

Which of the following are electromagnetic waves?

olchik [2.2K]

Answer:

Microwaves

Radiowaves.

5 0

2 years ago

Read 2 more answers