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2 years ago

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A 5 Kg bowling ball is thrown at a stationary 1.6 Kg bowling pin at 5 m/s. If the final velocity of the ball is 2.5 m/s. The fin

al velicoty of the bowling pin is _____m/s.

1 answer:

Sergio039 [100]2 years ago

7 0

Answer: hope this helps

Explanation:

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0

3 years ago

A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall

kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

$F = q(\vec v \times \vec B)$

so here we have

$F = qvB sin90$

since force is perpendicular to the velocity so here it must be centripetal force

here we have

$\frac{mv^2}{R} = qvB$

so we have

$R = \frac{mv}{qB}$

$R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}$

$R = 4.06 cm$

so here it will move in circle with radius 4.06 cm

8 0

2 years ago

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

kirill115 [55]

Answer:

The heat capacity of a sample is 37.7 J/K.

Explanation:

Given that,

Submerged temperature of tissue sample = 275 K

Mass of liquid nitrogen= 2 kg

Temperature = 70 K

Final temperature = 75 K

We need to calculate the heat

Using formula of heat

$Q=mc(T_{f}-T_{i})$

Put the value into the formula

$Q=2\times1.039\times10^{3}\times(75-70)$

$Q=10390\ J$

We need to calculate the heat capacity of a sample

Using formula of heat capacity

$\Delta S=\dfrac{Q}{T}$

Put the value into the formula

$\Delta S=\dfrac{10390}{275}$

$\Delta S=37.7\ J/K$

Hence, The heat capacity of a sample is 37.7 J/K.

7 0

3 years ago

Read 2 more answers

There is a 3-kg toy cart moving at 4 m/s. Calculate the kinetic<br> energy of the cart.

m_a_m_a [10]

The kinetic energy of the cart is 24 J.

<u>Explanation:</u>

The acceleration of a given mass from rest to the velocity is known as kinetic energy. It gains energy from acceleration and remains in this state until the speed of the object changes.

The kinetic energy is the given by,

K.E = 1/2 mv^2

Given the mass m = 3 kg, v = 4 m / s.

K.E = 1/2 $\times$ 3 $\times$ (4)^2

K.E = 24 J.

5 0

2 years ago

A proton initially at rest is accelerated by a uniform electric field. The proton moves 5.62 cm in 1.15 x 10^-6 s. Find the volt

krok68 [10]

Answer:

49.85 V

Explanation:

u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s

Let the electric field is E and voltage is V.

Use second equation of motion

s = ut + 1/2 a t^2

5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2

a = 8.5 x 10^10 m/s^2

m x a = q x E

E = m x a / q

E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)

E = 887.19 V/m

V = E x s

V = 887.19 x 5.62 x 10^-2 = 49.85 V

4 0

3 years ago