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3 years ago

What equation gives the position at a specific time for an object with constant acceleration?

Physics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

$x = x_{0}+v_{0}t+\frac{1}{2}at^{2}$ equation gives the position at a specific time for an object with constant acceleration

Explanation:

$x = x_{0}+v_{0}t+\frac{1}{2}at^{2}$ equation gives the position at specific time for an object having constant acceleration. Constant acceleration is referred as the change of velocity with respect to the time is known as the acceleration, but when the velocity changes occurs at constant rate this rate is termed as the constant acceleration. The constant acceleration can never be zero. The velocity changes but that change occurs in the consistently. The acceleration is affected by the mass.

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A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coi

weqwewe [10]

Answer:

The maximum emf generated in the coil is 60527.49 V

Explanation:

Given;

area of coil, A = 0.320 m²

angular frequency, f = 100 rev/s

magnetic field, B = 0.43 T

number of turns, N = 700 turns

The maximum emf generated in the coil is calculated as,

E = NBAω

where;

ω is the angular speed = 2πf

E = NBA(2πf)

Substitute in the given values and solve for E

E = 700 x 0.43 x 0.32 x 2π x 100

E = 60527.49 V

Therefore, the maximum emf generated in the coil is 60527.49 V

8 0

3 years ago

If you hold a bar magnet in each hand and bring your hands together, will the force be attractive or repulsive if the magnets ar

NemiM [27]

Answer:

Part A:

They will repel each other because same poles repel each other.

Part B)

They will attract each other because different poles attract each other..

5 0

3 years ago

Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A

kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × $\frac{ \lambda }{2}$ ........1

so λ = $\frac{2L}{10}$

and velocity is express as

V = $\sqrt{\frac{T}{\mu } }$ .................2

frequency for string A = f1 = $\frac{V1}{\lambda}$

f1 = $\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}$

f1 = $\frac{10}{2L} \sqrt{\frac{T1}{\mu } }$

and

f2 = $\frac{10}{2L} \sqrt{\frac{T2}{\mu } }$

beat frequency is = f2 - f1

put here value

beat frequency = $\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}$ - $\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }$

beat frequency = 13.87 Hz

6 0

3 years ago

Match the facts to the examples. The force of gravity increases with an increase in the mass of objects. Acceleration due to gra

sergeinik [125]

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8 0

3 years ago

Read 2 more answers

A person exerts a 19-N force on a cart attached to a spring and holds the cart steady. The cart is displaced 0.060 m from its eq

Mariana [72]

To solve this problem, apply the concepts related to Hooke's law. From there we will find the spring constant. Subsequently, applying Energy balance, which includes gravitational potential energy, elastic potential energy and kinetic energy, we will bury the system's energy. Finally, using the displacement expression for the simple harmonic movement, we will find the expression that describes the system.

PART A) The expression for the spring force is

$F=kx$

Here,

k = Spring constant

x = Displacement

Rearranging to find the spring constant we have that

$k = \frac{F}{x}$

$k = \frac{19}{0.06}$

$k = 316.66N/m \approx 320N/m$

PART B ) The gravitational potential energy acts on the spring holds the cart is zero. Since cart is placed in the equilibrium position. The kinetic energy of the cart is zero. Therefore the expression for the total energy is,

$E = (PE)_g+(PE)_{spring}+KE$