Answer:
3.43 m/s^2
Explanation:
Force is equal to mass times acceleration. (F=ma). You can use inverse operations to get the formula for acceleration, which is acceleration is equal to force divided by mass. (a=F/m). Since there are two forces here, the force friction (55 N), and the force applied (175 N), we must solve for the net force. To solve for the net force, you take the applied force (175 N) and subtract the frictional force from it (55 N). Thus, the net force is 120 N. With this done, we can now solve for our acceleration. 
Using the equation for acceleration, we take the force and divide it by mass.
120/35
Answer: 3.43* m/s^2**
*Note: This is rounded to the nearest hundredth, the full answer is: 3.42857143
**Note: In case you're confused, this is meters per second squared. 
 
        
             
        
        
        
After the ball reaches the top and begins its return back down, it's just a falling object that's been dropped.  
The amount that its downward speed increases every second is just the acceleration of gravity on Earth.
That's <em>9.81 meters per second</em> every second.
 
        
             
        
        
        
Answer: C) 0.25 m
Explanation: In order to explain this problem we have to consider the Faraday law, we have:
ε=-dФ/dt where ε the emf induced by the change of the magnetic field given by dФ/dt.
then ε=I*R=17*6=102 V
We have a coil then we have the magnetic flux as follow:
Ф=N*A*B then we have
dФ/dt= N*A*dB/dt  where A and N is the area and number of turn of the coil.
A=π*R^2 where R is the radius of teh coil.
Finally we have;
dФ/dt= N*π*R^2*dB/dt then
R= [dФ/dt/(N*π*dB/dt)]^1/2= [102/(180*π*3)]1/2=245.2*10^-3=≅0.25m
 
 
        
             
        
        
        
Is produced during cooler atmospheric conditions when a cold air mass moves across long expanses of warmer lake water. The lower layer of air, heated up by the lake water, picks up water vapor from the lake and rises up through the colder air above; the vapor then freezes and is deposited on the leeward (downwind) shores.[1]
The same effect also occurs over bodies of salt water, when it is termed ocean-effect or bay-effect snow. The effect is enhanced when the moving air mass is uplifted by the orographic influence of higher elevations on the downwind shores. This uplifting can produce narrow but very intense bands of precipitation, which deposit at a rate of many inches of snow each hour, often resulting in a large amount of total snowfall.
The areas affected by lake-effect snow are called snowbelts. These include areas east of the Great Lakes, the west coasts of northern Japan, the Kamchatka Peninsula in Russia, and areas near the Great Salt Lake, Black Sea, Caspian Sea, Baltic Sea, Adriatic Sea, and North Sea.
        
             
        
        
        
A particle has centripetal acceleration whenever it's a making a turn of radius R. If the particle is moving at a constant tangential speed v throughout the turn, then the magnitude of centripetal acceleration is
v²/R
If the particle is following a uniformly circular path, then it moves in a circle of radius R and travels a distance equal to its circumference, 2πR. Let T be the time it takes to complete one such loop. Then the entire circle is traversed with speed v = 2πR/T, so that the centripetal acceleration is also given by
v²/R = (2πR/T)²/R = 4π²R/T²