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4 years ago

How many millimeters are in .075cm ?

Physics

2 answers:

ser-zykov [4K]4 years ago

8 0

Answer: 0.75

Explanation:

take away the 0

FromTheMoon [43]4 years ago

7 0

0.75 millimeters :) you can solve this by multiplying the length value by 10. hope this helps

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L 3.2.3 Quiz: Friction

MAVERICK [17]

Answer:

the higher the value of the coefficient of friction the MORE the resistance to sliding .

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4 years ago

A 90kg cannon fires 100kg shell with muzzle velocity of 75m/s. Calculate the recoil velocity of the cannon relative to the groun

Gala2k [10]

Answer: 83.8 m/s

Explanation:

momentum of shell = momentum of cannon

M(s) × V(s) = M(c) × V(c)

M(s) = mass of shell, V(s) = velocity of shell

M(c) = mass of cannon, V(c) = velocity of cannon

100kg × 75m/s = 90kg × V(c)

7500 = 90 × V(c)

7500 ÷ 90 = V(c)

83.3 = V(c)

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3 years ago

A heavier mass m1 and a lighter mass m2 are 17.0 cm apart and experience a gravitational force of attraction that is 9.70 10-9 N

puteri [66]

Answer:

The value of m1 is 4.2 kg and value of m2 is 1 kg.

Explanation:

Note:- The value of force is assumed to be 9.70*10^-9 N.

The gravitational force F is given by the formula,

F=Gm1m2 / r^2

where F is the force, G is the universal gravitational constant, m1 and m2 are masses and r is the distance between the masses.

Then the product of the masses is,

m1m2= Fr^2/G

Given F=9.70*10^-9 N, r is 17.0 cm which is equal to 0.17 m.

Therefore product m1m2= 9.70*10^-9 N *(0.17 m)^2 / 6.67*10^(-11) N m^2 kg^-2

m1m2= 4.20 kg^2

m1=4.20 kg^2/m2

The sum of masses is m1+m2=5.20 kg

4.20/m2+m2=5.20

(m2)^2-5.20m2-4.20=0

After solving the above quadratic equation, mass m2 has values 4.2 kg and 1 kg.

By using m2=1 kg, m1 has greater mass 4.2 kg.

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2 years ago

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What are the characters associated with light as a wave

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Answer:

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all show the wave nature of light.

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3 years ago

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the eff

sashaice [31]

Complete Question

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2

Answer:

$h=1614m$