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Scrat [10]
3 years ago
8

If we assume that 60 % of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 m/s is transformed into heat tha

t flows into the nail and does not flow out, what is the temperature increase of an 8.00-g aluminum nail after it is struck ten times?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

45.6°C

Explanation:

Kinetic energy = \frac{1}{2}mv^2

Use m= 1.80kg and v=7.80m/s (mass and speed of hammer).

K = 0,5*1.80kg*(7.80m/s)^2 = K=54.8J

Heat is 60% of Kinetic energy. Q = 0.6*54.8J = 32.9J

As it is stuck 10 times the total heat is 10*32.9J = Total Heat = 329J

Use the equation Q = mC_v \Delta T to find change of temperature:

\Delta T = \frac{Q}{mC_v}

Q = 329J; m = 8.00g of aluminium; C_v = 0.900J/g°C (For aluminium)

\Delta T = \frac{329J}{8.00g*0.900J/g°C}

Calculating gives Change of Temperature = 45.6°C

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ikadub [295]

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency=\frac{velocity}{2 *length}

velocity =\sqrt{\frac{tension}{mass per unit length} }

mass per unit length=\frac{3.5}{1000*1.22}=0.00427\frac{kg}{m}

Now calculating velocity v=\sqrt{\frac{255}{0.00427} }

                                           =244.3\frac{m}{sec}

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = \frac{244.3}{2 *0.7} =174.5 hz

6 0
3 years ago
When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea
alekssr [168]

Answer:

The horizontal distance traveled by the projectile is 15.23 m.

Explanation:

Given;

angle of projection, θ = 25⁰

initial velocity of the projectile, u = 15 m/s

time of flight, t = 1.12 s

The the travelling path of the object is calculated as the range of the projectile

R = u_x t\\\\R = (15\ Cos \ 25^0) \times 1.12\\\\R = 13.595 \times 1.12\\\\R = 15.23 \ m

Therefore, the horizontal distance traveled by the projectile is 15.23 m.

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Answer:

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3 years ago
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What part of the hammer acts as the fulcrum when the hammer is used to remove a nail
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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
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