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Sloan [31]
3 years ago
11

Sound travels 2146 m through a material in 1.4 seconds. What is the material?

Physics
1 answer:
Sonbull [250]3 years ago
5 0
Your basically breaking the sound beerier   <span />
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A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s
ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = mu + 0 = mu

Momentum just after collision = (m + M)v

Therefore, mu=(M+m)v --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

f=\mu mg=-ma\\\\a=-\mu g

Now, using equation of motion, we can find the velocity just after collision.

0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}

Plug in the given values and find 'v'. This gives,

v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s

Now, using equation (1) and substituting the given values, we get:

0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

7 0
4 years ago
Homeostasis is necessary for these organisms to survive.
alexandr1967 [171]

Answer:

B, unicellular

Explanation:

8 0
3 years ago
What muscle movement do we use to close a joint.
marin [14]

Answer: Your using your skeletal muscle

Explanation:

3 0
3 years ago
CHEGG 42 mT magnetic field points due west. If a proton of kinetic energy 9 x 10-12 J enters this field in an upward direction,
alexdok [17]

Answer:

The  magnitude of the Force is  F = 697 *10^{-15}N  and the direction is South  

Explanation:

From the question we are told that

         The magnetic field point due west and since East point toward the positive x -axis(i)  then this magnetic field would be mathematically represented as

                \= B = 42(-i)mT = 42*10^{-3} (-i) T

Now from the question we are told that the kinetic energy is

             KE = 9*10^{-12}J

Now this kinetic energy can be mathematically represented as

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

Where m is the mass of proton which has a general value of

           m = 1.67*10^{-27}kg

Now making the subject of the formula

                v = \sqrt{\frac{KE}{0.5 * m} }

Substituting values we have

               v = \sqrt{\frac{9*10^{-12}}{0.5 * 1.67*10^{-27}} }

                 = 10.37*10^7m/s

Now from the question we are told that proton is moving upward which is in the positive z direction so the velocity of the proton would be in the positive

So the velocity would be

            \= v = 10.37*10^{7} \r k \ m/s

Now the magnetic Force can be mathematically represented as

          \= F = q \= v * \=  B

Where q is the charge on the proton which has a general value of  q =1.6*10^{-19}C

Now substituting the value

          \= F = 1.6*10^{-19 } * (10.37 *10^7) \r k * (42 *10^{-3})(-i)

              = 697*10^{-15} J

Now according to Fleming's left hand rule the direction of the magnetic force is south toward the negative Y - direction (-j)

So the force can be denoted as

                 \= F = 697*10^{-15}(-j) N

             

             

6 0
3 years ago
A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass
Talja [164]

Answer:

a)1.93 kg-m^2

b) 1.45  kg-m^2

c) = 0

d) 1.15 kg-m^2

Explanation:

mass of the bar M = 4 kg

length of the bar = 2 m

mass of balls m1= m2= 0.3 kg

moment of inertia of bar I= \frac{ML^2}{12}

about an axis perpendicular to the bar through its center.

a) MOI of bar + 2×m×(L/2)^2

I= \frac{ML^2}{12}+ 2m\frac{L}{2}^2

now putting the values of m, M and L as above and solving we get

I= 1.93 kg-m^2

b) perpendicular to the bar through one of the balls

I=M\frac{L^2}{3} +mL^2

I=4\frac{2^2}{3} +0.3\times4^2= 1.45  kg-m^2

c) parallel to the bar through both balls

zero as the no mass distribution along the parallel to the bar through both balls.

d) parallel to the bar and 0.500 m from it.

I=(M+2m_1)\frac{1}{2}^2

putting values and solving we get

1.15 kg-m^2

4 0
4 years ago
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