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EleoNora [17]
3 years ago
9

A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i

nstant when 28 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?
Physics
1 answer:
drek231 [11]3 years ago
8 0

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = \frac{28}{10} × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

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Answer:

B.  Speed = Wavelength x Frequency.Explanation:

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3 years ago
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A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block
kupik [55]

Answer:

a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is 89.289 joules.

e) 89.289 joules must be done to stop the bullet-block system.

f) The bullet-block system will travel 13.007 meters before stopping.

Explanation:

a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by  the initial momentum of the bullet:

p = m\cdot v_{o} (1)

Where:

p - Net momentum, in kilogram-meters per second.

m - Mass of the bullet, in kilograms.

v_{o} - Initial speed of the bullet, in meters per second.

If we know that m = 0.5\,kg and v_{o} = 50\,\frac{m}{s}, then the net momentum of the bullet-block system before the collision is:

p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)

p = 25\,\frac{kg\cdot m}{s}

The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The total energy of the bullet before the collision is its initial translational kinetic energy (K), in joules:

K = \frac{1}{2}\cdot m \cdot v_{o}^{2} (2)

K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}

K = 625\,J

The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:

v_{f} = \frac{m\cdot v_{o}}{m+M} (3)

Where:

v_{f} - Final speed, in meters per second.

M - Mass of the block, in kilograms.

If we know that m = 0.5\,kg, v_{o} = 50\,\frac{m}{s} and M = 3\,kg, then the final speed of the bullet-block system is:

v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)

v_{f} = 7.143\,\frac{m}{s}

The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is the translational kinetic energy of the system (K), in joules, is:

K = \frac{1}{2}\cdot (m + M)\cdot v_{f}^{2} (4)

K = \frac{1}{2}\cdot (0.5\,kg + 3\,kg)\cdot \left(7.143\,\frac{m}{s} \right)^{2}

K = 89.289\,J

The total energy of the bullet-block system after the collision is 89.289 joules.

e) By Work-Energy Theorem, magnitude of the work done by friction must be equal to the magnitude of the translational kinetic energy of the system. Hence, 89.289 joules must be done to stop the bullet-block system.

f) The maximum travelled distance of the bullet-block after the collision can be determined by means of Work-Energy Theorem and definition of work:

W = \mu_{k}\cdot (m+M)\cdot g\cdot s (5)

Where:

W - Work done by friction, in joules.

g - Gravitational acceleration, in meters per square second.

s - Travelled distance, in meters.

\mu_{k} - Kinetic coefficient of friction, no unit.

If we know that m = 0.5\,kg, M = 3\,kg, \mu_{k} = 0.2, g = 9.807\,\frac{m}{s^{2}} and W = 89.289\,J, then the travelled distance of the bullet-block system is:

s = \frac{W}{\mu_{k}\cdot (m+M)\cdot g}

s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

s = 13.007\,m

The bullet-block system will travel 13.007 meters before stopping.

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An iceberg of density 920kg/m^3 floats in seawater of density 1025kg/m^3 with volume of 10^6m^3.What is the total mass of the ic
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Im pretty sure the answer is 1.95x10^9
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An 80,000 kg plane is flying at 900 km/h at a height of 10 kilometers. What is its total mechanical energy? Group of answer choi
Romashka-Z-Leto [24]

The mechanical energy of the given flying object is 10,340 MJ. So the correct choice is 10,340 MJ.

Mechanical energy: The mechanical energy of an object is the sum of its potential energy and kinetic energy.

The formula for the mechanical energy E is:

E=K+U

where K is the kinetic energy and U is the potential energy.

Kinetic energy: The energy possessed by an object due to the motion.

The kinetic energy is given by the formula,

K=(1/2)*(mv^2)

where m is the mass of the object and v is the velocity of the object.

Potential energy: The energy possessed by an object due to its relative position from a given point.

The potential energy is given by the formula,

U=m*g*h

where g is the acceleration due to gravity and h is the height of the object from the ground.

Substitute K=(1/2)*(mv^2) and U=m*g*h in the formula of the mechanical energy to get the following formula,

E=(1/2)*(mv^2)+m*g*h

Given that m=80000 kg, h=10 km, and v=900 km/h, and the value of g is 9.8 m/s^(-2), the value of the total mechanical energy is,

E=(1/2)*(80,000 kg)(900 km/h )^2+(80,000 kg)*(9.8 m/s^(-2))*(10 km)

Note: 1 km= 1000 m and 1 km/h = 5/18 m/s. 1 MJ = 10^(6) J

E=(1/2)*(80,000 kg)*(900*(5/18) m/s )^2+(80000 kg)*(9.8 m/s^(-2))*(10000 m)

E=(1/2)*(80,000 kg)*(250 m/s)^2+7840*10^(6) J

E=2500*10^(6) J+7840*10^(6) J

E=10340*10^(6) J

E=10340 MJ

Learn more about Potential energy here:

brainly.com/question/15647581

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2 years ago
A cylinder has volume 50 cm and radius 4 cm find its volume. pls do it with the full process ​
CaHeK987 [17]

Answer:

Explanation:

What is the volume of a right cylinder with a circumference of 25π in and a height of 41.3 in? Explanation: The formula for the volume of a right cylinder is: V = A * h, where A is the area of the base, or πr2. Therefore, the total formula for the volume of the cylinder is: V = πr2h.

4 0
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