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2 years ago

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According to Newton's 1st law, the law of inertia, if you hit a baseball with a bat it will _____. A. hurt your hand from the re

action force B. accelerate according to the applied force C. fly in a straight line unless an outside force changes its course D. fall to the ground because of gravitational pull

2 answers:

solmaris [256]2 years ago

6 0

C. fly in a straight line unless an outside force changes its course.

Fofino [41]2 years ago

4 0

So the Newtons first law of motion states that an object in a state of uniform motion tends to remain in that states of motion unless an external force is applied and base on your question the possible answer would be letter B. accelerate according to the applied force

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There is a puncture in the inner tor of your bicycle. Give a method to detect the position of the puncture

Pani-rosa [81]

Answer:

Explanation:

The first method to engage is to listen to where the sound of air in the inner Tor escaping originated and look to see if u can find it. You can then feel the escape air with your hand.

You can Put it inside a container of water and see the bubble and rotate the inner tube to pass all of it through the water

7 0

3 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

2 years ago

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago

Describe the valence electrons in nitrogen and how it could bond to other atoms?

Gekata [30.6K]

Answer:

Each nitrogen molecule consists of two atoms of nitrogen that are bonded by a triple covalent bond. This is because each nitrogen atom has 5 valence electrons. Each atom can complete its octet by sharing three electrons.

I hope this helps!

4 0

2 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago