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Paladinen [302]
3 years ago
5

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

Physics
1 answer:
densk [106]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

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How many significant figures does 1700000000 have ?
MaRussiya [10]

Answer:

2 sig

Explanation:

1 and 7

3 0
3 years ago
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If a cross country runner covers a distance of 347 meters in 134 seconds what is her average speed?
a_sh-v [17]
347÷134=2.589552239 meters per second

2.589552239×60= 155.3731343 meters per hour

155.3731343 meters per hour= 0.096544389701642 miles per hour

hopefully this was right.
6 0
3 years ago
Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct
xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

7 0
2 years ago
A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0
2 years ago
Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to
Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength=\frac{velocity of sound }{frequency}

=\frac{343}{78}=4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

x-\left ( 8.2-x\right )=4.397    for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

x-\left ( 8.2-x\right )=0

2x=8.2

x=4.1m

Third

\left ( 8.2-x\right )-x=4.397

x=1.9015 m

6 0
3 years ago
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