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Tema [17]
3 years ago
11

Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

face, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood block move together as one object. How fast are they traveling? Please show your work
Physics
1 answer:
dybincka [34]3 years ago
8 0

The momentum of the wood and the bullet is equal to the momentum of the bullet before it hits the wood = 0.04 x 300 = 120 kg m/sHence (0.5 + 0.04)*v = 120 kg m/s, v = 120 kg m/s / 0.54 kg = 222.2 m/s3) Force = rate of change of momentum = 0.16(38+44)/0.002 = 6560 N

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A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates
tensa zangetsu [6.8K]

Answer:

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Explanation:

We have given the distance d = 1.45 cm = 0.0145 m

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There is a relation between potential difference and electric field

Electric field at a distance d due to a potential difference is given by

E=\frac{V}{d}, here E is electric field, V is potential difference and d is distance

So E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C

8 0
4 years ago
Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel
andrew-mc [135]

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

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\Delta W = m\cdot g\cdot \Delta z (1)

Donde:

m - Masa del montañero, medido en kilogramos.

g - Aceleración gravitacional, medida en metros por segundo al cuadrado.

\Delta z - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que m = 65\,kg, g = 9,807\,\frac{m}{s^{2}} y \Delta z = 500\,m, entonces el trabajo realizado por el montañero para subir es:

\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)

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7 0
3 years ago
What percent of all bolts fire within a cloud
Leni [432]

I think that it is 80%

5 0
3 years ago
A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r
Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

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Length of a rope, l =  4.56 m

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We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s

Let \omega is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0
3 years ago
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Leona [35]

Answer: Sound recording and production

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7 0
3 years ago
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