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yan [13]
4 years ago
15

We experience fictitious forces due to: a. Rotation of a reference frame b. Inertial reference frames c. Translational motion d.

Universal gravitation.
Physics
1 answer:
lisabon 2012 [21]4 years ago
8 0

Answer:

A.

Explanation:

A fictional force (also called force of inertia, pseudo-force, or force of d'Alembert, 5), is a force that appears when describing a movement with respect to a non-inertial reference system, and that therefore it does not correspond to a genuine force in the context of the description of the movement that Newton's laws are enunciated for inertial reference systems.

The forces of inertia are, therefore, corrective terms to the real forces, which ensure that the formalism of Newton's laws can be applied unchanged to phenomena described with respect to a non-inertial reference system. The correct answer is A.

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What is the speed of a 16 cm wave with a period of 8 seconds?
sp2606 [1]

The speed of a 16cm wave with a period of 8 seconds is 2cm/s

<h3>How to determine the speed</h3>

Using the formula;

Speed = distance ÷ time

Distance = 16cm

time = 8 seconds

Substitute into the formula

Speed = 16 ÷ 8

Speed = 2 cm/s

Therefore, the speed of a 16cm wave with a period of 8 seconds is 2cm/s

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3 0
2 years ago
where would information on the chemical and physical properties of a specific chemical be located in a laboratory or in the work
goldenfox [79]

Answer:

Both

Explanation:

8 0
3 years ago
Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

Angular speed of the football spiral, \omega=6.9\ rev/s=43.35\ rad/s

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
4 years ago
Two carts, cart AA (mass 4.00 kgkg) and cart BB (mass 6.50 kgkg) move on a frictionless, horizontal track. Initially, cart BB is
tankabanditka [31]

Answer:

V{_a}'=-0.95m/s

Explanation:

From the question we are told that:

Mass of cart A m_a=4kg

Mass of cart B m_a=6.50kg

Speed of cart AA V-{a}=4.00

Generally the equation for velocity of  A after collision V{_a}' is mathematically given by

V_{a}'=\frac{M_a-M_b}{M_a+M+b} V_a

V{_a}'=\frac{4-6.5}{4+M6.5} *4

V{_a}'=\frac{4-6.5}{4+M6.5} *4

V{_a}'=-0.95m/s

3 0
3 years ago
Determine the electrical force of attraction between two balloons with separate charges of +3.5 × 10-8 C and -2.9 × 10-8 C when
kenny6666 [7]

Answer:

The electrical force of attraction between the balloons is 2.16\times 10^{-5}\ N.

Explanation:

Given that,

Charge 1, q_1=+3.5\times 10^{-8}\ C

Charge 2, q_2=-2.9\times 10^{-8}\ C

Distance between the charges, d = 0.65 m

We need to find the electrical force of attraction between two balloons. It is given by the formula as :

F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{3.5\times 10^{-8}\times 2.9\times 10^{-8}}{(0.65)^2}\\\\F=2.16\times 10^{-5}\ N

So, the electrical force of attraction between the balloons is 2.16\times 10^{-5}\ N. Hence, this is the required solution.

3 0
3 years ago
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