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nikdorinn [45]
3 years ago
5

What is the Sl unit of Newton ?............​

Physics
2 answers:
Zielflug [23.3K]3 years ago
7 0

<em><u>N is the Sl unit of Newton. </u></em>

여기 당신의 대답이 있습니다 ..그것이 당신을 도울 수 있기를 바랍니다

kenny6666 [7]3 years ago
5 0

Answer:

<em>N</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>Newton</em>

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suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str
tatyana61 [14]

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

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3 years ago
A sport car moving at constant speed travels 110m in 5.0 s. if it then brakes and comes to a stop in 4.0 s, what is the magnitud
stiks02 [169]
D = 110 m,  t = 5 s
v o = 110 cs : 5 m = 22 m/s
-------------------------------------
v = v o - a t
v = 0 m/s,  v o =  22 m/s,  t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
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Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g. 
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3 years ago
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3 years ago
You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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2 years ago
What is serve in volleyball?
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