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butalik [34]
3 years ago
6

Eric has a mass of 60 kg. He is standing on a scale in an elevator that is accelerating downward at 1.7 m/s². What is the approx

imate reading on the scale?
a. 0 N
b. 490 N
c. 590 N
d. 690 N
Physics
1 answer:
vodomira [7]3 years ago
4 0

Answer:

option B

Explanation:

given,

mass of Eric = 60 Kg

acceleration o f the elevator downward (a)= 1.7 m/s²

acceleration due to gravity (g) = 9.8 m/s²

reading of scale = ?

when elevator is going down there will be a pseudo acting on Eric so, negative acceleration will be acting.

so, weight will be equal to

W_{apparent} - mg = - ma

W_{apparent} = m (g - a)

W_{apparent} = 60 ( 9.8 - 1.7 )

W_{apparent} = 60 x 8.1

W_{apparent} = 486 N

so, Weight will be approximately equal to 490 N

the correct answer is option B

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A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
Answer ?
morpeh [17]

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

5 0
2 years ago
Part of a neutralization reaction between an acid and a base is shown below. What is the second product of the reaction?
icang [17]
Salt compound. it is a double displacement
8 0
3 years ago
Read 2 more answers
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blagie [28]

Answer:

  3, 4, 2

Explanation:

I can't see the final ones.

4 0
3 years ago
A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 0
3 years ago
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