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3 years ago

What is it asking MathPhys im sorry i dont know but i tried teaching myself physics but i cant

Physics

1 answer:

RSB [31]3 years ago

6 0

Answer:

The puck B remains at the point of collision.

Explanation:

This is an elastic collision, so both momentum and energy are conserved.

The mass of both pucks is m.

The velocity of puck B before the collision is vb.

The velocity of puck A and B after the collision is va' and vb', respectively.

Momentum before = momentum after

m vb = m vb' + m va'

vb = vb' + va'

Energy before = energy after

½ m vb² = ½ m vb'² + ½ m va'²

vb² = vb'² + va'²

Substituting:

(vb' + va')² = vb'² + va'²

vb'² + 2 va' vb' + va'² = vb'² + va'²

2 va' vb' = 0

va' vb' = 0

We know that va' isn't 0, so:

vb' = 0

The puck B remains at the point of collision.

You might be interested in

BaLLatris [955]

Answer:

The conduction path or simply the wires connected between different components in a circuit.

Explanation:

The wire makes up the path for the electricity to flow and most of the electricity flows through this. It is like a road connecting two house or buildings in a town and the traffic of vehicles is the electricity (current).

5 0

3 years ago

A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth

Alex73 [517]

Answer:

$F_{Earth}$ = 15.57 N

$F_{Moon}$ = 2.60 N

$F_{Uranus}$ = 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

$F_{Earth}$ =m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

$F_{Moon}$ =m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

$F_{Uranus}$ =m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0

2 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago

• Describe the main ideas of physical science

arlik [135]

Branch of science that studies matter and its motion through space and time, along with related concepts such as energy and force. Describing the nature, measuring and quantifying of bodies and their motion, and dynamics.

5 0

3 years ago

An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w

adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 × $10^{-2}$ m

diameter d = 0.074 cm = 74 × $10^{-5}$ m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 × $10^{-5}$ ×5 × $10^{-2}$

area = 1162.3892 × $10^{-5}$ m²

so here power emitted is express as

power emitted = E × σ × area × (temperature)^4

put here all value

power emitted = 0.300× 5.67 × 1162.3892 × $10^{-5}$ × (3068)^4

power emitted = 1.75 W

5 0

3 years ago