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ch4aika [34]
3 years ago
15

Acceleration can be defined as the rate at which velocity ________

Physics
2 answers:
Savatey [412]3 years ago
8 0

Answer:

The speed is 20 m/s, and the direction is "downward". Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.

Explanation:

EleoNora [17]3 years ago
7 0

Acceleration is the rate of the change in velocity.

The answer would be the rate at which velocity changes.

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The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coeffi
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Answer:

Explanation:

Let T be the tension in the cord.

Impulse by cord = change in momentum of block A .

T x 5s = 10 ( 2 -0) = 20

T = 4 poundal .

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Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .

= 8 ( 32 + .4 ) = 259.2 poundal

Frictional force on block A = 259.2 - 4 = 255.2 poundal

μ x 10 x 32 = 255.2

320μ = 255.2

μ =0 .8 .

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A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the
iris [78.8K]

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>
  • Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
  • n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
  • Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

  • So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

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What is the genotype of the male?
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A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant
AlekseyPX

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

2\alpha \theta = \omega_f^2 -\omega_I^2

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}

<u>ωf = 8.8 rad/s</u>

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

<u>v = 2.2 m/s</u>

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3 years ago
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