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3 years ago

Shawn and his bike have a total mass of

1 answer:

5 0

Speed = distance / time
Speed = 700m / 792s = 0.88383838m/s

0.5 times mass times velocity^2
0.5 times 52.9 times 0.8..^2
= 20.6619541

Not sure if it is actually correct tho

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In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o

ioda

Answer:

Explanation:

We shall apply law of conservation of momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1 )

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

2 )

Initial kinetic energy of the system

= 1/2 x 0.1 x 4²

= 0.8 J

Final kinetic energy of the system = 7.3 x 10⁻²

Loss of energy = .8 - .073

= .727 J

This energy was converted into internal energy of the system .

3 )

increase in entropy = dQ / T

Here dQ = .727 J

T = 300 ( Constant )

dQ / T = 2.42 X 10⁻³ J/K

6 0

3 years ago

A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a d

Snowcat [4.5K]

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

$U = 0.5 C V^2$

The capacitance is given by

$C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J$

7 0

2 years ago

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would

natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a) $v_{es} =v_{rms}$ = 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given, $v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

7 0

3 years ago

As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force

fgiga [73]

Answer: $107.8\ J$

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

$\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J$