How do the temperature curves in your graph show the range of temperature?

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

2 years ago

A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p

Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s$

It takes 2.84403 seconds to reach the highest point

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m$

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

$s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s$

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

6 0

3 years ago

Why is gasoline so flammable

Karo-lina-s [1.5K]

Flammable and combustible liquids themselves do not burn. It is the mixture of their vapours and air that burns. Gasoline, with a flashpoint of -40°C (-40°F), is a flammable liquid. Even at temperatures as low as -40°C (-40°F), it gives off enough vapour to form a burnable mixture in air.

Hope this helps

I looked this up

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8 0

3 years ago

Read 2 more answers

A boy exerts an unknown horizontal force as he pulls a 52 N sled across packed snow. The coefficient of friction is 0.12. If a p

Tems11 [23]

Answer:

is there a pic?

Explanation:

7 0

2 years ago

Read 2 more answers

A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s

AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0

3 years ago