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skelet666 [1.2K]
3 years ago
13

Fusion is a type of _____.

Physics
1 answer:
Tresset [83]3 years ago
7 0
Fusion is the type of nuclear reaction. It powers the Sun.
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What is the mechanical advantage of a nail puller where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the oth
Mandarinka [93]

Answer:

Explanation:

Mechanical Advantage is the ratio of the distance of the input load (Li)from the pivot to the output load applied to the pivot(Lo)

MA = Li/Le

Given;

Li = 45cm

Lo = 1.8cm

MA = 45/1.8

MA = 25

Hence the mechanical advantage is 25

Also MA is expressed in terms of the force ratio which is the ratio of the Load to the effort applied.

MA = Load/Effort

Given

Load = 1250N

MA = 25

Effort = ?

Substitute

25 = 1250/Effort

Effort = 1250/25

Effort = 50N

Hence the minimum force exerted on the load is 50N

3 0
3 years ago
What is the magnitude of the force that lifts a 3-kilogram object straight upwards?
frez [133]

Any force of 29.4 Newtons or greater can do it.

4 0
3 years ago
Read 2 more answers
Quartz has an index of refraction of 1.46. Diamond has in index of refraction of 2.42. In which material does a light ray enteri
Ksivusya [100]
<span>Diamond slowdown light more than Quartz , because diamonds have a greater index of refraction. Light will bend when its move from one medium to another. The Index of Refraction of Material is found by comparing the speed of light in their respective mediums.</span>
4 0
3 years ago
Read 2 more answers
A rural mail carrier logged the mileage on the odometer at the start of the mail route as 31,500 miles. After 6.5 hours, the car
dusya [7]

Answer:32.77\ mi/hr

Explanation:

Given

Initially Reading on the odometer is 31,500\ miles

Final reading on the odometer is 31,713\ miles

Time taken is t=6.5\ hr

average velocity =\dfrac{\text{Displacement}}{\text{time}}

v_{avg}=\dfrac{31713-31500}{6.5}

v_{avg}=\dfrac{213}{6.5}

v_{avg}=32.77\ mi/hr

Thus the average velocity of mail truck is 32.77\ mi/hr

8 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
2 years ago
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