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Sauron [17]
4 years ago
9

What is one way you could reduce the friction between two surfaces

Physics
1 answer:
Vlad [161]4 years ago
6 0
You could put ice, put it on wheels!
You might be interested in
How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m
vovikov84 [41]
<h2>Answer: 1000 J</h2>

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:  

1J=(1N)(1m)=Nm  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance d to overcome the resistance of the weight of the box (its 200 N).  

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work W performed is the product of the force exerted to push the box F=50N by the distance traveled d. as shown in equation (1).

Hence:

W=(50N)(20m)  

W=1000Nm=1000J >>>>This is the work  

8 0
3 years ago
New attempt is in progress. Some of the new entries may impact the last attempt grading. Incorrect. The hammer throw is a track-
Morgarella [4.7K]

This question involves the concepts of centripetal force, range of projectile and projectile motion.

The magnitude of centripetal force is "2812.8 N".

First, we will find the velocity of the ball by using the formula of the range of the projectile.

R = \frac{v^2Sin2\theta}{g}

where,

R = range of projectile = 86.75 m

v = speed = ?

θ = launch angle = 47.9°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

86.75\ m = \frac{(v)^2Sin(2)(47.9^o)}{9.81\ m/s^2}\\\\v = \sqrt{\frac{(86.75\ m)(9.81\ m/s^2)}{Sin95.8^o}}

v = 29.25 m/s

Now, we will use the formula to find out the centripetal force:

F_c = \frac{mv^2}{r}

where,

F_c = Centripetal Force = ?

m = mass of the ball = 7.3 kg

v = speed = 29.25 m/s

r = radius = 2.22 m

Therefore,

F_c = \frac{(7.3\ kg)(29.25\ m/s)^2}{2.22\ m}

<u>Fc = 2812.8 N = 2.812 KN</u>

<u />

Learn more about centripetal force here:

brainly.com/question/11324711?referrer=searchResults

5 0
3 years ago
A 3.0 kg ball is lifted two meters upward in two seconds, and a 6.0 kg ball is lifted one meter upward in one second? Which requ
Sav [38]

Answer:

1. They both uses same energy

2. The 6 kg ball requires more power than 3kg ball

Explanation:

Sample 1

m = 3kg

g= 10m/s^2

h = 2m

t = 2secs

W = mgh = 3 x 10 x 2 = 60J

P= w/t = 60/2 = 30watts

Sample 2

m = 6kg

g= 10m/s^2

h = 1m

t = 1sec

W = mgh = 6 x 10 x 1 = 60J

P= w/t = 60/1 = 60watts

They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball

3 0
4 years ago
I need help please :)
Masja [62]

Answer:

20

Explanation:

AB, AC, AD, AE, BC, BD, BE, ....

4 0
3 years ago
Explain why during a cold winter, the air temperatures are generally higher on snowy days than on clear days.
Aleks04 [339]
I don’t know that is weird
4 0
4 years ago
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