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Nikolay [14]
3 years ago
15

Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they

apply a force of 43 newtons, and the crate just starts moving. If the coefficient of static friction is 0.11, what is the value of the normal force?
Physics
2 answers:
dangina [55]3 years ago
8 0
Answer is 3.9 x 10^2
maw [93]3 years ago
7 0
 <span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is: 

Ff = UsN 

where Us is the coefficient of static friction and N is the normal force. 

In order to get the crate moving you must first apply enough force to overcome the static friction: 

Fapplied = Ff 

Since Fapplied = 43 Newtons: 

Fapplied = Ff = 43 = UsN 

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11 

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
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Answer:

the boat would be deeped by 3200 m

Explanation:

Given that

The boat arrives back after 4 seconds

And, the speed of the sound in water is 1,600 m/s

We need to find out how much deep is the water

So,

As we know that

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Here we divided by 2 because the boat arrives back

= (1600 × 4) ÷ 2

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
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shusha [124]

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Explanation:

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