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ValentinkaMS [17]
3 years ago
11

A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s

. If one third of the hammer’s kinetic energy is converted to the internal energy of the hammer and spike, how much does the total internal energy increase? Answer in units of J
Physics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

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Apply SUVAT

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Which statement describes something that should always be done at the end of a calculation?
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C) Check that the numerical answer is reasonable and the units are what you expected.

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A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red
Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }       ......(1)

Here \lambda_{s} is source wavelength, \lambda_{o} is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, \lambda_{s} = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, \lambda_{0} = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }

\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84

{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056

1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}

\frac{v}{c}=\frac{0.2944}{8.056}

v = 0.036c=0.036\times3\times10^{8}

v = 1.08 x 10⁷ m/s  

8 0
3 years ago
Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from
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Electric field between the plates of parallel plate capacitor is given as

E = \frac{Q}{A\epsilon_0}

here area of plates of capacitor is given as

A = 0.055 * 0.055

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also the maximum field strength is given as

E = 3 * 10^6 N/C

now we will plug in all data to find the maximum possible charge on capacitor plates

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3 years ago
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An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

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