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Maksim231197 [3]
3 years ago
7

A height of 6 ft 3 in is equal to _______. (Round to 3 significant digits)

Physics
2 answers:
aksik [14]3 years ago
7 0

Answer:

option c

Explanation:

1 feet = 12 inch

1 inch = 2.54 cm

o, 6 ft 3 in = 6 x 12 + 3 = 75 inch

75 inch = 75 x 2.54 cm = 190.5 cm

So, the answer is 191 cm

Olin [163]3 years ago
6 0

A height of 6 ft 3 in is equal to _______. (Round to 3 significant digits)  

A - 0.191 cm

B - 19.1 cm

<u>C - 191 cm</u>

6 ft 3 in = 190.5 cm

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3 years ago
The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?​
Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0
2 years ago
Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A c
Andru [333]

Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

Explanation:

Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂

x = F/k = F/k₁ + F/k₂

1/k = 1/k₁ + 1/k₂

k = (1/k₁ + 1/k₂)⁻¹

B

The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂ + x₃

x = F/k = F/k₁ + F/k₂ + F/k₃

1/k = 1/k₁ + 1/k₂ + 1/k₃

k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

8 0
2 years ago
Read 2 more answers
A car has a mass of 1,200 kg. What is its acceleration when the engine exerts a force of 600 N? (Formula: F=ma)
Rasek [7]

Answer:

Option (a)

Explanation:

Given that,

Mass of a car, m = 1200 kg

Force exerted by the engine, F = 600 N

Noe force,F = ma

a is the acceleration of the engine

a=\dfrac{F}{m}\\\\a=\dfrac{600\ N}{1200\ kg}\\\\a=0.5\ m/s^2

So, the acceleration of the car is 0.5 m/s².

6 0
3 years ago
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