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Artemon [7]
3 years ago
5

A tree limb of mass 12 kg falls straight down. If air resistance exerts 27 N of force on the limb as it falls, what is the net f

orce on the tree limb
Physics
2 answers:
Mandarinka [93]3 years ago
8 0

Answer: 91N down

Explanation:

Karolina [17]3 years ago
6 0

The net force is 27 n

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What is the resultant force of 500g on abject accelerating at 5m/s2
max2010maxim [7]

Answer: 2.5 N

Explanation:

m = 500g = 0.5Kg

a = 5m/s2

F = ma = 0.5 x 5 = 2.5 N

6 0
3 years ago
A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What
ladessa [460]

Answer:

New pressure is 0.534 atm

Explanation:

Given:

Initial volume of the gas, V₁ = 250 mL

Initial pressure of the gas, P₁ = 1.00 atm

Initial temperature of the gas, T₁ = 20° C = 293 K

Final volume of the gas, V₂ = 500 mL

Final pressure of the gas = P₂

Final temperature of the gas, T₁ = 40° C = 313 K

now,

we know for a gas

PV = nRT

where,

n is the moles

R is the ideal gas constant

also, for a constant gas

we have

(P₁V₁/T₁) = (P₂V₂/T₂)

on substituting the values in the above equation, we get

(1.00 × 250)/293 = (P₂ × 500)/313

or

P₂ = 0.534 atm

Hence, the <u>new pressure is 0.534 atm</u>

5 0
3 years ago
If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
A cheetah starts from rest and accelerated at 8.7 m/s^2 for 3s. How far did the cheetah go in that time
nika2105 [10]

Answer:

cheetah goes 52.2 m in that time.

Explanation:

4 0
3 years ago
When external forces acting on an object are balanced, what will happen to the object's
Brilliant_brown [7]

Answer:

The objects morion will remain the same

Explanation:

8 0
3 years ago
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