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Brilliant_brown [7]
3 years ago
13

A hockey puck is hit on a frozen lake and starts moving with a speed of 13.60 m/s. Exactly 6.2 s later, its speed is 7.20 m/s. (

a) What is the puck's average acceleration?
Physics
1 answer:
stellarik [79]3 years ago
3 0

Answer:

-1.03 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².

Mathematically, acceleration is expressed as

a = (v-u)/t ........................ Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t  = time.

Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.

Substituting into equation 2

a = (7.20-13.60)/6.2

a = -6.4/6.2

a = -1.03 m/s²

Note: a is negative because, the hockey puck is decelerating.

Hence the average acceleration = -1.03 m/s²

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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
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Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

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            DE = 13.606 (1/1 - 1/3²)

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            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

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3 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

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M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

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h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

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I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

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