Ask question

Ask question

All categories

3 years ago

13

a ball rolls off a platform that is 5.0m above the ground with a horizontal speed of 6.0 m/s how far from the base of the platfo

rm does the ball hit the ground?

1 answer:

svp [43]3 years ago

7 0

Please elaborate more on your question so I can help you

You might be interested in

The magnetising current in a transformer is rich in ______________ harmonic?

sukhopar [10]

The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher.

5 0

3 years ago

A force of attraction would exist between

Kryger [21]

Answer:

D. two positively charged objects

3 0

3 years ago

A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of

bulgar [2K]

m = mass of the car moving in horizontal circle = 1750 kg

v = Constant speed of the car moving in the horizontal circle = 15 m/s

r = radius of the horizontal circular track traced by the car = 45.0 m

F = magnitude of the centripetal force acting on the car

To move in a circle . centripetal force is required which is given as

F = m v²/r

inserting the above values in the formula

F = (1750) (15)²/(45)

F = (1750) (225)/(45)

F = 1750 x 5

F = 8750 N

6 0

3 years ago

Find the unknown mass of the block 1 needed to balance the bar. Assume that the mass of the bar is negligible. Block 1 is locate

Natasha2012 [34]

Answer:

Incomplete question: The masses of the blocks m₂ = 1.5 kg and m₃ = 2 kg

Explanation:

Given data:

L₁ = length = 0.85 m

L₂ = 0.25 m

L₃ = 0.5 m

m₂ = 1.5 kg

m₃ = 2 kg

Question: Find the unknown mass of the block 1 needed to balance the bar, m₁ = ?

The torque is zero (intermediate point of the bar)

$-m_{1} gL_{1} +m_{2} gL_{2} +m_{3} gL_{3} =0$

Is negative because mass 1 is to the left of the coordinate system (see the diagram)

$m_{1} =\frac{m_{2} gL_{2}+m_{3} gL_{3} }{L_{1} } =\frac{(1.5*0.25)+(2*0.5)}{0.85} =1.6176kg$

4 0

4 years ago

what is the net force of a 6 kg object that goes from a velocity of 0.5 m/s to a velocity of 3 m/s over the course of 1.5 second

baherus [9]

Answer:

∑F = 10.2 N

Explanation:

We have:

Initial velocity: 0.5 m/s

Final velocity: 3 m/s

Time: 1.5 s

We have all of the components needed to calculate acceleration. Let's do that, shall we?

a = vf-vo/t

a = 2.5/1.5

a = 1.7 $m$ / $s^{2}$

Now, look at the Net Force equation:

∑F = ma

Plug in the variables, to get:

∑F = (6)(1.7)

∑F = 10.2 N (You can round this according to significant digits)

3 0

3 years ago