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2 years ago

A sled is slowing down at the bottom of a snowy hill.

Physics

1 answer:

Kitty [74]2 years ago

5 0

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

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7 0

3 years ago

A 25 kg box on a horizontal frictionless surface is moving to the right at a speed of 5.0 m s. The box hits and remains attached

seropon [69]

Answer:

Explanation:

Work done by the spring is negative .

Work done by force F creating displacement d is given by the following expression .

Work = F x d

Both force and displacement are vector quantity .

When direction of force and direction of displacement is same , work is positive . When direction of force and direction of displacement is opposite , work is negative .

When spring is compressed , it exerts a restoring or opposing force in a direction opposite to the direction of displacement of box . Hence here force is opposite to displacement . Restoring force acts opposite to displacement . Hence work done by spring on box is negative .

6 0

2 years ago

Spaceship 1 and spaceship 2 have equal masses of 200 kg. Spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 10 m/s

m_a_m_a [10]

Answer:

2000 kg m/s

Explanation:

The momentum of an object is a vector quantity whose magnitude is given by

$p=mv$

where

m is the mass of the object

v is the velocity of the object

and its direction is the same as the velocity.

In this problem, we have:

- Spaceship 1 has

m = 200 kg (mass)

v = 0 m/s (zero velocity)

So its momentum is

$p_1 =(200)(0)=0$

- Spaceship 2 has

m = 200 kg (mass)

v = 10 m/s (velocity)

So its momentum is

$p_2=(200)(10)=2000 kg m/s$

Therefore, the combined momentum of the two spaceships is

$p=p_1+p_2=0+2000=2000 kg m/s$

3 0

2 years ago

Jane is a team leader. Match her leadership and teamwork skills to the appropriate descriptions.making her team understand the r

gtnhenbr [62]

Jane is motivating and empowering her team.

7 0

2 years ago

1) If I have 10 A of current running through my series circuit (which has 5

timofeeve [1]

Answer:10 A

Explanation:

According to the KCL (Kirchhoff's current laws)

8 0

2 years ago