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steposvetlana [31]
4 years ago
11

A rake acts as a machine by _____ over which the force is applied.

Physics
1 answer:
puteri [66]4 years ago
4 0
A rake acts as a machine by decreasing the applied force and decreasing the distance over which the force is applied. A machine is defined as an object that reduces the input work needed to make something function. Basically, it is something that makes our manual labor easier. Since Work is a function of force and distance, a machine decreases the force and distance needed.
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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

5 0
3 years ago
A fan cart with the fan set to high rolled across the floor. The cart's speeds with the fan on high are shown below. If the fan
Vadim26 [7]

Answer:

Speed of cart's might be less than the high speed after 5 seconds.

Explanation:

Given that,

A fan cart with the fan set to high rolled across the floor.

Let the speed of fan cart with set to high is x\ cm per second.

The fan supplies a force to the cart. If a lower fan speed were used, less force would be applied. This would cause a slower change in the cart's speed. So, the cart would be rolling more slowly than x\ cm per second after 5 seconds. The speed of cart's might be less than x\ cm per second.

Force is needed

A. for a moving object to keep moving at the same speed and direction

B. for a moving object to change its speed

C. for a motionless object to remain still

D. to prevent a moving object from turning

Hence,

Speed of cart's might be less than the high speed after 5 seconds.

3 0
3 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0
2 years ago
A student pushes on a 5 kg box with a force of 20 N forward. The force of sliding friction
mart [117]

Answer:

Explanation:

F=20N

mu=-10N

F=ma

20-10=10

10=(5)(a)

a=2 m/s^2

4 0
3 years ago
Liquid can flow but solid cannot give reason ​
Airida [17]

<em>Answer:</em>

<em>well..</em>

<em>Explana</em><em>tion</em><em>:</em>

<em>L</em><em>iquid</em><em> can flow but solid cannot because of differences in their properties</em>

<em>property of liquid which lets it flow:</em>

  • <em>i</em><em>nter-particular</em><em> space is large</em>
  • <em>inter-particular attraction is small</em><em> </em><em>t</em><em>hese</em><em> properties tend to make the molecules of liquid free to flow</em><em> </em>

<em>property</em><em> </em><em>of</em><em> </em><em>solid</em><em> </em><em>which</em><em> </em><em>tends</em><em> </em><em>to</em><em> </em><em>obstruct</em><em> </em><em>flow</em><em>:</em>

  • <em>inter-particular</em><em> </em><em>spa</em><em>c</em><em>e</em><em> </em><em>is</em><em> </em><em>small</em><em> </em><em>and</em><em> </em><em>so</em><em> </em><em>it's </em><em>compac</em><em>t</em>
  • <em>inter-molecular</em><em> </em><em>attra</em><em>ction</em><em> </em><em>is</em><em> </em><em>strong</em><em> </em><em>hence</em><em> </em><em>no</em><em> </em><em>tenden</em><em>cy</em><em> </em><em>to</em><em> </em><em>flow</em>

<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>this</em><em> </em><em>helps</em><em>!</em>

6 0
3 years ago
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