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LenKa [72]
2 years ago
11

A 6kg object speeds up from 5 m/s to 20 m/s. find p

Physics
2 answers:
kicyunya [14]2 years ago
6 0

as we know tha momentum is product of mass and velocityi.e p=mv but here in this question velocity changes from 5m/s to 20 m/s so here formula becomes

p=(final velocity -initial velocity )mass

so (20-5)6=90kgm/s ANS

mrs_skeptik [129]2 years ago
4 0

Momentum = mass x speed

The object's momentum is 30 kg-m/s at the beginning, 120 kg-m/s at the end, and its change in momentum is 90 kg-m/s .

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mass 1 = 1.60kg

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Radioactive isotopes are used in medicine to create images of internal
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Answer:

The correct option is D) Fission

Explanation:

There are several methods through which Radioactive isotopes are created.

  1. Using a nuclear reactor that has a field of neutrons, insert a stable sample such as Lutetium-176. When it gets bombarded with neutrons, it acquires some, and fission is said to have occurred. Note that when Lutetium-176 acquires a neutron, it becomes radioactive Lu-177.
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3 0
3 years ago
What type of weather would a continental Polar air mass bring
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Answer:

Continental polar ( cp):

Explanation:

Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.

6 0
3 years ago
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

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7 0
2 years ago
Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0
3 years ago
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