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makkiz [27]
3 years ago
15

a wall switch is connected to a wall outlet. A lamp is plugged into the same outlet. You turn the switch on, but the lamp stays

off. Give three reasons why this might happen.
Physics
1 answer:
jolli1 [7]3 years ago
5 0
1:   only half  the outlet is switched  and the lamp is in the  other half
2:  the lamp is turned off.
3:   The  light bulb  is burned out
4:  the switch might be broken
5:   the fuse   might be blown
6: the electricity might be off
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You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
3 years ago
A plane is flying horizontally with speed
Kay [80]

Answer:

2. ahead of the package.

Explanation:

Using y' - y = ut - 1/2gt², we find the time, t it takes the package to hit the ground. So, u = initial vertical velocity of package = 0 m/s, y = initial position of package = 3970 m, y' = final position of package = 0 m, g = acceleration due to gravity = 9.8 m/s².

Substituting the variables into the equation, we have

y' - y = ut - 1/2gt²

0 m - 3970 m = 0t - 1/2 × (9.8 m/s²)t²

-3970 m = -(4.9m/s²)t²

t² = -3970 m ÷ -4.9 m/s²

t² = 810.2 s²

t = √810.2 s²

t = 28.5 s

Using v = u' + at, we find the horizontal acceleration of the plane. Since the initial horizontal velocity of the package is that of the plane, u' = 162 m/s, v = 0 m/s since the package stops and t = 28.5 s when the package stops.

So, a = (v - u')/t

a = (0 m/s - 162 m/s)/28.5 s

a = -162 m/s/28.5 s

a = -5.68 m/s²

Using v² = u'² + 2as, we find the horizontal distance ,s where the package stops.

So, s =  (v² - u'²)/2a

Substituting the values of the variables, we have

s =  ((0 m/s)² - (162 m/s)²)/(2 × -5.68 m/s²)

= - 162 m²/s²/(-11.36 m/s²)

= 14.26 m

The horizontal distance d the plane moves after releasing the package is d = u't = 162 m/s × 28.5 s = 4617 m

Since d = 4617 m > s = 14.26 m, the plane would be ahead of the package when the package hits the ground.

6 0
3 years ago
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