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xz_007 [3.2K]
3 years ago
13

Two forces are applied to a 2.0 kg block on a frictionless horizontal surface. F1 = 8.ON is applied to the left while F2 = 3.0 N

is applied to the right. What is the
acceleration of the block?


A.) 2.5 m/s^2 to the left

B.) 1.5 m/s^2 to the right

C.) 4.0 m/s^2 to the left

D.) 2.5 m/s^2 to the right
Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

It is C 5.0 m/s 2 to the right

Explanation:

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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

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<h2>_____________________________________ </h2><h2>Question 8: </h2>

\Large\textbf{Diode:}  

A diode is a device that allows current to flow in only one direction.

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8 0
3 years ago
A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

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8 0
2 years ago
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Answer:

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Explanation:

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2 years ago
1. Driving at slower speeds than traffic flow _____________ .
r-ruslan [8.4K]

<u>The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance d due to a static charge having charge q is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}

Here, E is the electric field intensity, q is the amount of charge and r is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

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<u> </u>

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords:  Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

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