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skad [1K]
3 years ago
12

A 50 kg ball traveling at 20 m/s would have______ kinetic energy

Physics
1 answer:
ANTONII [103]3 years ago
6 0

Explanation:

Data:-m=50kg ,v=20m/sec , K.E=? solution;-formula K.E=1/2mv² ,K.E=1/2×50×(20)² ,K.E=25×400=10000J

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Which statements about acceleration are true?
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C is the correct answer, hope it helps
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Show that the energy of a magnetic dipole m in a magnetic field B is U--m B
Juli2301 [7.4K]

Answer:

showm

Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field \vec{B} then the torque exerted by the field on the dipole is

\tau = m\times B

\tau=mBsin\alpha

Now to rotate the dipole in the field to its final position the work required to be done is

U=\int \tau d\alpha

U=\int mBsin\alpha d\alpha

U= -mBcos\alpha

U=-\vec{m\times \vec{B}}

Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

5 0
3 years ago
How is matter changed from one state of matter to another? *
Digiron [165]

Explanation:

Matter is changed from one state to another by addition or removal of heat and suitable pressure conditions.

When a solid is heated, it normally melts and changes to liquids which on heating changes to vapor. The randomness of the particles increases from solid to liquid state and to gaseous states.

Also, a gas can be cooled to liquid and on further cooling transformed into a solid matter.

These phase changes are brought about by energy changes in a system. Some form of matter can also sublime by changing form solid to gas and vice versa.

4 0
3 years ago
When melting of a metamorphic rock occurs, it changes into what?
Thepotemich [5.8K]

Answer:

The upper limit of metamorphism occurs at the pressure and temperature of wet partial melting of the rock in question. Once melting begins, the process changes to an igneous process rather than a metamorphic process. During metamorphism the protolith undergoes changes in texture of the rock and the mineral make up of the rock.

Explanation:

i hope this helps

8 0
3 years ago
A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The
qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0
3 years ago
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