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Korvikt [17]
3 years ago
11

A temperature of 20°c is equivalent to approximately

Physics
2 answers:
Yanka [14]3 years ago
8 0
C would be your answer
Nastasia [14]3 years ago
3 0
The answer is c 68 degree
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The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
3 years ago
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HELP! FOR A TEST!
Sati [7]

Answer:

<h3>On a velocity vs time graph the slope of the line represents the acceleration of the object. With a slope of zero, the object is moving at a constant velocity in the positive (+) direction during this five minute interval. ... Displacement and distance can both be determine on a velocity vs.</h3>

4 0
3 years ago
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A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?
Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h  

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3  

V = 6.307 x 10^-5 m^3  

By density = m/V  

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

6 0
3 years ago
We cannot consider something to be work unless
iren [92.7K]
<span>force applied causes movement of an object in the same direction as the applied force.</span>
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3 years ago
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You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j
Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

E= \frac{1}{2} kx^2

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

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6 0
1 year ago
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