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3 years ago

A body is under going non uniform circular motion work done by tangential force on body is

Physics

1 answer:

dybincka [34]3 years ago

8 0

Answer:

The tangential force will act as a torque on the body, increasing its angular velocity and thus also increasing its kinetic energy. By the work-kinetic-energy theorem, work has been done on the body. Yes, in non-uniform circular motion the work done on the object is non-zero, for the reason you stated.

Explanation:

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describe what happens when light strikes a translucent material, transparent material , opaque material . give one specific exam

CaHeK987 [17]

Answer: according to https://www.doe.virginia.gov When light strikes translucent materials, only some of the light passes through them. The light does not pass directly through the materials. ... Opaque objects block light from traveling through them. Most of the light is either reflected by the object or absorbed and converted to thermal energy.

Explanation:

6 0

3 years ago

A circular loop of wire 75 mm in radius carries a current of 113 A. Find the (a) magnetic field strength and (b) energy density

Roman55 [17]

The magnetic field strength is 9.47 ×10⁻⁴ T

The energy density at the center of the loop is 0.36 J/m³

<h3>Calculating Magnetic field strength & Energy density </h3>

From the question, we are to find the magnetic field strength

The magnetic field strength of a loop can be calculated by using the formula,

$B = \frac{\mu_{0} I}{2R}$

Where B is the magnetic field strength

$\mu_{0}$ is the permeability of free space $(\mu_{0}=4\pi \times 10^{-7} \ N/A^{2})$

$I$ is the current

and R is the radius

From the give information,

$R = 75 \ mm= 75 \times 10^{-3} \ m$

and $I = 113 \ A$

Putting the parameters into the formula, we get

$B = \frac{4\pi \times 10^{-7} \times 113}{2 \times 75 \times 10^{-3} }$

$B = 9.47 \times 10^{-4} \ T$

Hence, the magnetic field strength is 9.47 ×10⁻⁴ T

Now, for the energy density

Energy density can be calculated by using the formula,

$u_{B} = \frac{B^{2} }{2\mu_{0} }$

Where $u_{B}$ is the energy density

Then,

$u_{B}= \frac{(9.47\times 10^{-4} )^{2} }{2 \times 4\pi \times 10^{-7} }$

$u_{B} = 0.36 \ J/m^{3}$

Hence, the energy density at the center of the loop is 0.36 J/m³

Learn more on Magnetic field stregth & Energy density here: brainly.com/question/13035557

7 0

2 years ago

A 770 n man stands in the middle of a frozen pond of radius 10.0 m. he is unable to get to the other side because of a lack of f

oksano4ka [1.4K]

Momentum is conserved throughout this scenario.

Before the man does anything, the total momentum of him and his book is zero. So we know that it'll be zero after he throws the book.

Momentum = (mass) x (velocity)

The man gives the book (1.2 kg)x(10 m/s north) = 12 kg-m/s north
of momentum.

Since the total momentum must be zero, the man himself picks up 120 kg-m/s of momentum south.

(his mass)x(his v) = 120 kg-m/s south = (770 kg-m/s^2/9.8 m/s^2)x(V).

His velocity southward = (120 x 9.8) / (770) m/s .

He needs to reach the shore 10m away.

Time = distance/speed

= (10 x 770) / (120 x 9.8) seconds

= 6.55 seconds

3 0

3 years ago

A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration

bazaltina [42]

Answer:

Stress = F / A force per unit area

A = 3.00 cm^2 = 3 E-4 m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied

F/3 = 2.4E4 N if force not to exceed limit (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g

3 0

3 years ago

A double-slit experiment yields an interference pattern due to the path length difference from light traveling through one slit

Lapatulllka [165]

Answer:

Therefore the correct statement is B.

Explanation:

In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.

The sum of all these waves results in the diffraction curve of the slit that has the shape

I = Io sin² θ /θ²

Where the angle is a function of the wavelength and the width of the slit.

From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)

Therefore the correct statement is B

6 0

3 years ago

A body is under going non uniform circular motion work done by tangential force on body is​

A body is under going non uniform circular motion work done by tangential force on body is