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4 years ago

What is relation of acceleration of a body to its mass and applied force

1 answer:

3 0

The law states that external forces cause objects to accelerate, and the amount of acceleration is directly proportional to the net force and inversely proportional to the mass of the object.

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A cylindrical region of radius R contains a uniform magnetic field parallel to its axis. The field is zero outside the cylinder.

PtichkaEL [24]

The magnitude of the induced electric field is (RdB/dt)/4

The induced electric field is gotten from

-∫E.dl = dФ/dt where E = induced electric field, dl = path length vector, Ф = magnetic flux through cylindrical region = AB where A = area of magnetic flux = πR² where R = radius of cylindrical region and B = magnetic field.

So, -∫E.dl = dФ/dt

-∫E.dl = dAB/dt

-∫Edlcos0 = AdB/dt (where E.dl = Edlcos0 = Edl since E and dl are parallel to each other.)

So -∫Edl = πR²dB/dt

-E∫dl = πR²dB/dt (∫dl = 2πr since the integral is the circumference of the path)

-E(2πr) = πR²dB/dt (we integrate dl from r = 0 to 2R)

-E2π(2R - 0) = πR²dB/dt

-E4πR= πR²dB/dt

E = πR²dB/dt ÷ 4πR

E = -(RdB/dt)/4

So, the magnitude of the induced electric field is (RdB/dt)/4

Learn more about induced electric field here:

brainly.com/question/15730392

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2 years ago

27. Improvements in space technology have greatly helped scientists better understand the stars, planets, and other

stellarik [79]

Answer:

Explanation:

Technological advancements in astronomy have led to the development of thousands of products such as

new materials, medical devices, and communications satellites.

For example, the production of instruments like telescopes, microscopes, and other measuring devices now enables us to identify things that we couldn't examine with the naked eye.

Similarly, the development of communication satellites has led to the emergence and improvements in the mobile network industry.

8 0

3 years ago

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression $x=12 cm$

Potential energy Stored in spring $=72 J$

Suppose k is the spring constant of spring

Potential Energy of spring is given by $=\frac{kx^2}{2}$

$\frac{k(0.12)^2}{2}=72$

$k(0.12)^2=144$

$k=10,000 N/m$

$k=10 kN/m$

for 100 J energy

$\frac{k(x_0)^2}{2}=100$

$10\times 10^3\cdot (x_0)^2=200$

$(x_0)^2=2\times 10^{-2}$

$x_0=0.1414$

$x_0=14.14 cm$

6 0

3 years ago

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s

skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0

4 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$