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Studentka2010 [4]
3 years ago
9

How does friction affect speed?

Physics
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Friction is a force tends to oppose relative motion between bodies when they are in direct contact. Remember this friction does not oppose motion, it opposes relative motion.

Explanation:

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How much does a 45 kg rock weigh
bazaltina [42]

Answer:

441.45N

Explanation:

Weight = mass(kg) x acceleration (m/s^2)

Weight = 45kg x 9.81m/s^2=441.45N

Recall that weight =mass of the rock x gravitational force acting on the rock

Therefore, the rock will weigh = 441.45N

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3 years ago
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Answer:

I think deposition of C is the oldest deposit

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Write down the address of Earth in as much detail as possible
jeka57 [31]
Position #3
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5 0
3 years ago
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
kumpel [21]
<h3>Answer : 360J</h3>

<h3>Way to do : </h3>

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

<h3>A bit of explanation : </h3>

W = Work (J)

F = Force / weight (N)

s = distance (m)

6 0
3 years ago
A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b
guapka [62]
<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=200m/s is the bullet's initial speed

\theta=0 because we are told the bullet is shot horizontally

t is the time since the bullet is shot until it hits the ground

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=1.5m  is the initial height of the bullet

y=0  is the final height of the bullet (when it finally hits the ground)

g=9.8m/s^{2}  is the acceleration due gravity

<h2>Part (a):</h2>

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}   (3)

0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}   (4)

Finding t:

t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}   (5)

Then we have the time elapsed before the bullet hits the ground:

t=0.553s   (6)

<h2>Part (b):</h2>

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

x=V_{o}cos\theta t   (1)

Substituting the knonw values and the value of t found in (6):

x=200m/s.cos(0)(0.553s)   (7)

x=200m/s(0.553s)   (8)

Finally:

x=110.656m  

4 0
4 years ago
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