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Shalnov [3]
3 years ago
10

If someone feels intent on fighting with you, what should you do to prevent getting into a fight?

Physics
1 answer:
Arturiano [62]3 years ago
6 0

Answer:

avoid verbal argument and walk away

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A player strikes a hockey puck giving it a velocity of 30.252 m/s. The puck slides across the ice for 0.267 s after which time i
erica [24]

Answer:

The average drag force is  1.206 (-i)  N

Explanation:

You have to apply the equations of<em> Impulse</em>:

I=FmedΔt

Where I and Fmed (the average force) are vectors.

The Impulse can also be expressed as the change in the <em>quantity of motion</em> (vector P)

I=P2-P1

P=mV (m is the mass and v is the velocity)

You can calculate the quantity of motion at the beggining and at the end of the given time:

Replace the mass in kg, dividing the mass by 1000 to convert it from g to kg.

P1=(0.179kg)(30.252m/s) i=  5.414 i kg.m/s

P2=0.179kg)(28.452m/s) i = 5.092 i kg. m/s

Where i is the unit vector in the x-direction.

Therefore:

I= 5.092 i - 5.414 i = -0.322 i

The average drag force is:

Fmed= I/Δt = -0.322 i/ 0.267s = -1.206 i N

3 0
4 years ago
16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both
Daniel [21]

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

p=mv=(95)(8.2)=779 kg m/s

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

\Delta p = p' -p =0-779  = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

p_f + p_t = p'

where

p_f is the initial momentum of the fullback

p_t is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

p_f = 779 kg m/s\\p' = 0

Therefore, we can find the tackle's original momentum:

p_t = p'-p_f = 0-(779) = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

p=-779 kg m/s is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s

So, he was moving at 6.1 m/s in the direction opposite to the fullback.

4 0
4 years ago
In 1666 at the age of 23, what scientist
scoray [572]

Answer:

<h3>B - Isaac Newton</h3>

Explanation:

He first thought of his system of gravitation which he hit upon by observing an apple fall from a tree,

The incident occurring in the late summer of 1666.

3 0
3 years ago
A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0°. What is the va
zloy xaker [14]

Answer:

<em>T</em><em>he value of work being done on the object is 958J.</em>

Explanation:

<em>Work</em><em> done</em><em> </em><em>is </em><em>equal</em><em> to</em><em> </em><em>force</em><em> </em><em>multiply</em><em> by</em><em> </em><em>distance,</em><em> </em><em>but </em><em>when</em><em> </em><em>the </em><em>angle</em><em> </em><em>is </em><em>between</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>distance</em><em> </em><em>work</em><em> </em><em>done=</em><em>Force</em><em> (</em><em>cos</em><em> </em><em>theta)</em><em> </em><em>×</em><em> </em><em>distance</em>

<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>Force(</em><em>cos </em><em>theta</em><em>)</em><em> </em><em>×</em><em> </em><em>distance</em>

<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>2</em><em>5</em><em>N</em><em>(</em><em>cos </em><em>4</em><em>0</em><em>.</em><em>0</em><em>°</em><em>)</em><em> </em><em>×</em><em> </em><em>5</em><em>0</em><em>m</em>

<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>2</em><em>5</em><em>N</em><em>(</em><em>0</em><em>.</em><em>7</em><em>6</em><em>6</em><em>0</em><em>)</em><em> </em><em>×</em><em> </em><em>5</em><em>0</em><em>m</em>

<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>19.15N </em><em>×</em><em> </em><em>5</em><em>0</em><em>m</em>

<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>957.5J </em><em>=</em><em> </em><em>9</em><em>5</em><em>8</em><em>J</em>

<em>T</em><em>herefore</em><em> the</em><em> </em><em>value</em><em> of</em><em> </em><em>work</em><em> </em><em>being</em><em> </em><em>done </em><em>on </em><em>the</em><em> </em><em>object</em><em> </em><em>is </em><em>9</em><em>5</em><em>8</em><em>J</em><em>.</em>

3 0
2 years ago
Identify the type of heat transfer occurring when chocolate melts in your hand
qwelly [4]
Conductive maybe? I forgot but I’m pretty sure it’s conducive
3 0
4 years ago
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