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Igoryamba
3 years ago
15

23.

Physics
1 answer:
Volgvan3 years ago
4 0

Answer:

1 m = 100 cm....so 2.5 m = (2.5 * 100) = 250 cm

a = 1st shelf

b = 2nd

c = 3rd

d = 4th

a + b + c + d = 250

b = 2a + 18

c = a - 12

d = a + 4

a + (2a + 18) + (a - 12) + (a + 4) = 250

5a + 10 = 250

5a = 250 - 10

5a = 240

a = 240/5

a = 48 cm <== 1st shelf

b = 2a + 18 = 2(48) + 18 = 114 cm <== 2nd shelf

c = a - 12 = 48 - 12 = 36 cm <== 3rd shelf

d = a + 4 = 48 + 4 = 52 cm <== 4th shelf

so 2nd shelf is 114 cm

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Encuentre en km.h la velocidad de un tigre que corre 550 km.h en 69min​
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v = 478.26 km/h

Explanation:

The question is "find in km.h the speed of a tiger that runs 550 km in 69min"

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v=\dfrac{550\ km}{1.15\ h}\\\\v=478.26\ km/h

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
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Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

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3 years ago
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