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HACTEHA [7]
3 years ago
7

Give one example where friction is a hindrance. ​

Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

an example of when friction is a hindrance, is if someone with shorts is sliding down a slide. their skin rubbing on the slide will cause friction and will slow them down. another example is if you're dragging a large wooden box across the floor. the wood and the floor interacting cause friction and will make it more difficult to push the box.

Explanation:

hope this helps :D

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In an electrical circuit, which of the following controls whether or not electrons flow through the circuit?
love history [14]
A SWITCH. IT controls the flow of electrons in a circuit

4 0
4 years ago
You add 500ml of water at 10celsius to a 100ml of water at 70 Celsius. What is the most likely final temperature of the mixture?
Schach [20]

I wanna say it’s C.

3 0
4 years ago
Read 2 more answers
What acceleration will you give to a 25kg box if you push it with a force of 85N
Grace [21]
<h3>Answer:</h3>

3.4 m/s²

<h3>Explanation:</h3>

We are given;

  • Mass of the box as 25 kg
  • Force is 85 N

We are required to determine the acceleration;

  • According to second newton's law of motion force is given by the product of mass and acceleration.
  • That is;

Force = ma

Rearranging the formula;

a = F ÷ m

Therefore;

acceleration = 85 N ÷ 25 kg

                       = 3.4 m/s²

Thus, the acceleration of the box will be 3.4 m/s²

5 0
3 years ago
Can someone help meeeeee... show how to solve it plzzzzzzzz
liubo4ka [24]
<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have a gravitation force F_{1}=16units, given by the formula written at the beginning. Let’s rename the distance r as d:

F_{1}=G\frac{m_{1}m_{2}}{d^2}     (1)

And we are asked to find the gravitation force F_{2} with a given distance of \frac{d}{2}:

F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}      

F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}     (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate G m_{1}m_{2} in both equations:

From (1):

Gm_{1}m_{2}=F_{1}{d}^{2}     (3)

From (2):

Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}     (4)

If (3)=(4):

F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}     (5)

Now we have to find F_{2}:

F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}      

F_{2}=4F_{1}     (6)

If F_{1}=16 units:

F_{2}=(4)(16 units)        

F_{2}=64 units>>>>This is the new force of attraction     

3 0
3 years ago
on earth a block is placed on a frictionless table. When a 50 North horizontal force is applied to the block, it accelerates at
melamori03 [73]

As per Newton's II law we know that

F = ma

here

F = applied unbalanced force

m = mass of object

a = acceleration of object

now it is given that force F = 50 N North applied on block on earth due to which block will accelerate by 4 m/s^2

so here from above equation

50 = m* 4

m = \frac{50}{4} = 12.5 kg

Now we took another situation where block is placed on surface of moon and again force F = 25 N is applied on the block

So we will again use Newton's II law

F = ma

25 = 12.5 * a

a = \frac{25}{12.5}

a = 2 m/s^2

so block will accelerate on moon by acceleration 2 m/s^2

5 0
3 years ago
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