How many hydrogen atoms could bind with oxygen

An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the gr

amm1812

Answer:

q=3.5*10^-4

Explanation:

concept:

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

The tension force:

when the plane is uncharged

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

when the plane is charged

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k) √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

5 0

3 years ago

A campus bird spots a member of an opposing football team in an amusement park. The football player is on a ride where he goes a

kenny6666 [7]

Answer:

No..

Explanation:

As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.

As the person is on circular arc constantly moving it will never meet that drop.

6 0

4 years ago

. Which of the following is NOT a property of water? A. Ability to dissolve many substances B. Constant volume upon freezing C.

ss7ja [257]

Answer:

a

Explanation:

water is only able to desolve two substances , salt , sugar,

7 0

3 years ago

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Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic

Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by $\frac{1}{2}kx^{2}$ where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, $\frac{1}{2}k(x1)^{2}$

⇒ $\frac{1}{2}k(\frac{w}{k})^{2}$

⇒ $\frac{w^{2}}{2k}$

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, $\frac{1}{2}2k(x2)^{2}$

⇒ $\frac{1}{2}2k(\frac{w}{2k})^{2}$

⇒ $\frac{w^{2}}{4k}$

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is $\frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=$ 2:1

4 0

3 years ago

At the equator earth rotates with a velocity of about 465 m/s.

Dafna1 [17]

The given velocity is 465 m/s.

Part a.
$465 \, \frac{m}{s} =(465 \times 10^{-3} \, \frac{km}{s})*( 3600 \, \frac{s}{h} ) = 1674 \, \frac{km}{h}$
Answer: 1674 km/h

Part b.
$1674 \frac{km}{h} = (1674 \, \frac{km}{h})*(24 \, \frac{h}{day} ) = 40176 \, \frac{km}{day}$
Answer: 40,176 km/day.

3 0

3 years ago

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