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STALIN [3.7K]
3 years ago
5

Is Throwing a baseball and a football Newton's 2nd law of motion ?

Physics
2 answers:
ella [17]3 years ago
7 0

Answer:

Yes!

Explanation:

Good luck!

prohojiy [21]3 years ago
5 0

Answer:

yes

Explanation:

if I'm wrong them I sorry but I'm sertain it is

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Da 6.0 kg wooden crate slides across a wooden floor
IrinaK [193]

Answer:coe?

Explanation:

4 0
2 years ago
Steve just completed a woodworking project. now he wants to put a fine finish on his work
vesna_86 [32]
Pad sander I think I’m not positive tho
8 0
3 years ago
What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ
madreJ [45]

Answer:

L=4.19*10^{-6}H

Explanation:

The self-inductance of a solenoid is defined as:

L=\frac{\mu N^2A}{l}

Here \mu_0 is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:

L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H

4 0
3 years ago
In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of t
tino4ka555 [31]

The force required to pull the two hemispheres is 46622.72N

<h3>Calculation and Parameters</h3>

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12)

= 928 millibars.

(928 x 100)

= 92,800N/m^2.

Therefore, the required force to pull the two hemispheres is

(92800 x 0.5024)

= 46622.72N.

Read more about force here:

brainly.com/question/15300777

#SPJ1

6 0
2 years ago
I just don't know how to do the question (a) and (b)​
ozzi

Using the equation

F=ma

we can observe that you have to apply a non-zero net force to an object in order to make it accelerate. In fact, if the net force is zero you have

0=ma \iff a=0

Since we're assuming n\neq 0

Now, if the 12N force is applied, the object moves with a constant speed. A constant speed means no acceleration, since by definition the acceleration is a change in speed.

If this sounds counterintuitive to you (why I'm applying a force but I have to acceleration?) think of when we drive a car: even if you want to keep your speed constant, you still have to use the gas pedal, just enough so that the push of the motor balances exactly the road/wheels friction. If you give less gas, the friction becomes stronger, and the car slows down. If you give more gas, the motor push becomes stronger, and the car accelerates.

Back to your exercise: constant speed means to acceleration, so the net force must be zero. This implies that the friction force is exactly 12N.

If the force is increased to 18N, there will be a net force of 6N pushing the object, causing it to accelerate. Using again the same equation of before, and plugging the 3kg mass in the equation, we have

F=ma \iff 6=3a \iff a=2

So, the object moves with constant acceleration and initial speed of 10m/s for 0.2 seconds. It's final speed will be

v = v_0+at = 10+2\cdot 0.2 = 10.4

5 0
3 years ago
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