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I am Lyosha [343]

4 years ago

7

In 8.5 s a fisherman winds 2.4 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

. The line is being reeled in at a constant speed. Determine the angular speed of the reel.

1 answer:

SSSSS [86.1K]4 years ago

4 0

Answer:

9.412 rad/s.

Explanation:

Velocity is the rate of change of an object's position.

V = x/t

Where x is the distance in m

= 2.4 m

t is time taken in s

= 8.5 s

V = 2.4/8.5

= 0.2824 m/s.

Equating linear velocity and angular velocity,

V = ω*r

Where,

ω Is the angular speed in rad/s

r is the radius of the circle in m

= 3 cm

= 3cm * 1m/100 cm = 0.03 m

ω = V/r

= 0.2824/0.03

= 9.412 rad/s.

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

3 years ago

a sample contains 36g of a radioactive isotope. how much radioactive isotope remains in the sample after 3 half-lives

barxatty [35]

After 1st half life it will be 36/2 i.e. 18gm then 9 gm and then 4.5 grams simply
Hope this is what you were searching for!

8 0

3 years ago

Read 2 more answers

A space shuttle orbits Earth at a speed of 21,000 km/hr. How far does it go in 3.5 hrs?

inn [45]

Since it travels at 21,000 kilometers per hour, you'd just multiply that with 3.5 to get 73,500. So your answer is 73,500.

5 0

4 years ago

A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change

lions [1.4K]

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

now the mass of gas is given as

$m = n M$

$m = (0.06)(28)$

$m = 1.65 g$

now density of gas when its volume is increased to 3.3 L

so we will have

$\rho = \frac{m}{V}$

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$

5 0

3 years ago

A ball is thrown horizontally from the top of a building and strikes the sidewalk after 2.8 s. How tall is the building? Neglect

Brut [27]

Answer:

h = 38.41 m

Explanation:

Given that,

A ball is thrown horizontally from the top of a building and strikes the sidewalk after 2.8 s.

We need to find the height of the building. Let it is h.

Initial speed of the ball, u = 0

Using second equation of motion to find h as follows :

$h=ut+\dfrac{1}{2}at^2\\\\\text{Put u = 0 and a = g}\\\\h=\dfrac{1}{2}gt^2\\\\=\dfrac{1}{2}\times 9.8\times 2.8^2\\\\=38.41\ m$

So, the building is 38.41 m tall.

3 0

3 years ago