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3 years ago

6

A baseball player hits a baseball with a bat. The mass of the ball is 0.05 kilograms. The ball accelerates at 2,000 meters per s

econd squared.
What is the net force, to the nearest newton, that accelerates the ball?

1 answer:

AlexFokin [52]3 years ago

8 0

Answer:

550

Explanation:

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1:a boy 2 a girl pulling a heavy crate at the same time w/10 unite of force each.What is the net force acting on the object

vovangra [49]

The maximum magnitude of the net force on the box is 20 N, which is only possible if the boy and the girl pull the box together in the same direction, horizontally and parallel to the ground.
The minimum magnitude of the net force on the box is 0 N, which will occur when the boy and the girl pull the box together in the parallel but opposite direction.
If either of them pulls at an angle from the horizontal, then the magnitude of the net force will be between 0 N and 20 N.

8 0

3 years ago

Guys please help me on the rest of the numbers

wel

Answer:

C. 100

D.3

E. 33.3

Explanation:

C. Mechanical Advantage=Load / Effort

= 200N

--------

100N

Therefore,. = 100

D. I. Velocity Ratio= distance moved by the effort / distance moved by load

= 30cm/10cm

= 3

II. Efficiency= M.A / V.R

= 100/3

= 33.33

8 0

3 years ago

The answer is not B.

mezya [45]

Answer:

D

Explanation:

15m/ s

i hope this helps

4 0

2 years ago

Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you

dybincka [34]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.

So,

F = $\frac{Kq1q2}{r^{2} }$

Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.

Let's suppose, q1 = q

So,

if q1 = q

then

q2 = Q-q

Product of Charges = q1 x q2

Now, it is:

Product of Charges = q x (Q-q)

So,

Product of Charges = qQ - $q^{2}$

And the expression qQ - $q^{2}$ is clearly a quadratic expression. And clearly its roots are 0 and Q.

So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.

So, the midpoint is:

q = $\frac{Q + 0}{2}$

q = Q/2 and it is the highest value of each charge in order to get the greatest force.

8 0

3 years ago

We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength

san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

$\theta=1.22\frac{\lambda}{b}$

θ = 50.0*10^-7 rad

λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

$b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm$

hence, the smallest diameter objective lens is 13.4cm

8 0

3 years ago

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