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kvasek [131]
3 years ago
12

A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.

012. What is the magnitude of the frictional force?
Physics
1 answer:
pogonyaev3 years ago
3 0

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

F=\mu N

where

\mu is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is \mu=0.012. The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so

N=65 N+52 N=117 N

Therefore, the frictional force is

F=\mu N=(0.012)(117 N)=1.40 N

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Aleksandr-060686 [28]

Answer:0.302

Explanation:

Given

mass of crate m=27 kg

Force required to set crate in motion is 80 N

Once the crate is set  in motion 56 N is require to move it with constant velocity

i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force

thus  

f_s(static\ friction)=\mu \cdot N

where \muis the coefficient of static friction and N is Normal reaction

N=mg

f_s=\mu mg

\mu mg=80

\mu =\frac{80}{27\times 9.8}

\mu =0.302

7 0
4 years ago
How can people help stop erosion on beach
seraphim [82]

Answer:

<h3>since erosion is unavoidable the problem becomes discovering ways to prevent it. present beach erosion prevention methods include sand bags,vegetation,seawalls,sand dunes,and sand fences.</h3>
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3 years ago
James yap run fast while dribbling the ball as he crosses the middle part of the basketball court he changes his phase to avoid
Delicious77 [7]

Answer:

The answer is "Speed".

Explanation:

In the question, the Speed is the correct answer because it can be viewed as its rate from, which a length is covered via an object. It is also divided by time into other components of distance. Its SI unit of speed seems to be the meter/second, but still, the kilometer/hour or, in the Us UK, miles per hour is the most basic fundamental of speed in everyday use.

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4 years ago
A duck swimming on the surface of a pond has an
Gemiola [76]

From the given information:

  • Taking the movement of the Duck in the North as the x-direction
  • The movement of the Duck in the East direction as the y-direction

However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.

<h3>In vector form;</h3>

The initial velocity is:

\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}

The acceleration is:

\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}

The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;

Then, it implies that time (t) =  4 seconds.

Using the first equation of motion:

v^{\to} = u ^{\to} + a^{\to} t

Then, we can replace their values into the equation of motion in order to determine the speed:

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}

Collect like terms:

\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0   )\hat x+(  0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}

\mathbf{v^{\to} =0.87500   \hat x- 1.01629 \hat y}

Thus, the magnitude is:

\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}

\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}

\mathbf{v^{\to} =\sqrt{1.79848}}

\mathbf{v^{\to} =1.34 \ m/s}

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

Learn more about vectors here:

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