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Delicious77 [7]
3 years ago
13

What should you do when a crate is too heavy to be lifted by the pulley?

Physics
1 answer:
kenny6666 [7]3 years ago
7 0
Use a wedge, I would say. It would depend on the answers.
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Assume that the Styrofoam slab and the fur are both initially neutral, and that the slab charged negatively after it is rubbed w
BARSIC [14]

Charging phenomenon is basically transfer of electrons from one system to other. When a system loses the electrons the it becomes positively charged and when a system gains electrons then it will become negatively charged

so here if slab is rub against fur then one of them will lose the electrons while other will gain the electrons and hence one will get negatively charged and other will get positive charge

Here it is given that slab get negatively charged so correct options are

<em>3. Electrons move from the fur to the slab. </em>

<em>6. The fur becomes positive after rubbing the slab. </em>

4 0
3 years ago
As a 5.0 × 10^2-newton basketball player jumps
AysviL [449]
     This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
     Using the Newton's Secound Law, we have:

F_{R}=ma \\ W+E-N=ma \\ N =W+E \\ N=5*10^2+10*10^2 \\ \boxed {N=1.5*10^3N}

Number 3

If you notice any mistake in my english, please let me know, because i am not native.
4 0
3 years ago
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.
drek231 [11]

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under  gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I  am  sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

8 0
3 years ago
What is 16.558 m/s rounded to three significant figures?
In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0
3 years ago
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is
AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

3 0
3 years ago
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