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viktelen [127]
3 years ago
8

Explain your observation in terms of force and Newtons 3rd law

Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A

Explanation:

the law of force show that focuses will always come in 2 or pairs and it also says that one body cannot exert a force on another without experiencing a force itself.

( hope that helps :) #pk1208 is here to help )

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ivanzaharov [21]
The answer would be 46.482 because you multiply 18.3 by 2.54 because for every inch you get 2.54 centimeters
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Which change indicates that the universe is expanding?
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Red shift of distant galaxies
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A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14
Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

Ef - Ei = 0

(K+U)final - (K+U)initial =0

(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

((1/2)vB² + g*hB = (1/2 )vA²+ g*hA

(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

v_{B} = \sqrt{237.16}

vB = 15.4 m/s : speed of the cart at B

4 0
3 years ago
When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

Now,

The angular velocity can be calculated as:

v = \omega R

Thus

\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s

8 0
3 years ago
The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is
77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

\% Percent = \frac{V_{nucleus}}{V_{atom}}*100

\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100

\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100

\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100

\% Percent = 7.51*10^{-13}*100

\% Percent = 7.51*10^{-11}\%

Therefore the percent of the atom's volume is occupied by mass is 7.51*10^{-11}\%

3 0
3 years ago
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