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3 years ago

Which of the following forces acting upon an atom are the strongest?

Physics

1 answer:

iren [92.7K]3 years ago

3 0

Answer:

Nuclear Forces

Explanation:

Because strong nuclear forces work best within shorter distance.

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A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.

Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light $=15 feet$

height of man $=6 feet$

$\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s$

From diagram

$\frac{15}{y}=\frac{6}{y-x}$

$5(y-x)=2y$

$3y=5x$

differentiate both sides

$3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Tip of shadow is moving at the rate of

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s$

(b)rate at which length of his shadow is changing

Length of shadow is $y-x$

differentiating w.r.t time

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s$

7 0

3 years ago

A Or b <br> Question in picture

Helen [10]

Answer:

the answer to the question is indeed B

5 0

3 years ago

Read 2 more answers

Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max

yanalaym [24]

Answer:

$\theta=39.49^{\circ}$

Explanation:

Maximum height of the pumpkin, $H_{max}=9.99\ m$

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

$H_{max}=\dfrac{v^2\ sin^2\theta}{2g}$

On rearranging the above equation, to find the angle as :

$\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})$

$\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})$

$\theta=39.49^{\circ}$

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0

3 years ago

A 10 N force is used to push a 30 kg box across the floor, moving it a distance of 20 m. There is no friction. What is the magni

tekilochka [14]

Work=Force • Distance
W= 10N • 20m
W= 200 J

5 0

3 years ago

Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g

Vladimir79 [104]

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

5 0

3 years ago