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Bingel [31]
3 years ago
13

A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude

0.50\,\dfrac{\text m}{\text s^2}0.50 s 2 m ​ 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. After 2.0\,\text{s}2.0s2, point, 0, start text, s, end text, the bee is moving to the right with a speed of 2.75\,\dfrac{\text m}{\text s}2.75 s m ​ 2, point, 75, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. What was the velocity of the bumblebee right before the breeze?
Physics
1 answer:
lesya692 [45]3 years ago
7 0

Answer:

3.75 m/s

Explanation:

Given:

v = 2.75 m/s

a = -0.50 m/s²

t = 2.0 s

Find: v₀

v = at + v₀

2.75 m/s = (-0.50 m/s²) (2.0 s) + v₀

v₀ = 3.75 m/s

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The photographer realizes that with the lens she is currently using, she can't fit the entire landscape she is trying to photogr
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She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

What is focal length?

The focal length is a measure of how strongly the system converges or diverges light.

A positive focal length indicates that a system converges light, while a negative focal length indicates that the system diverges light.

For a standard rectilinear lens,

FOV = 2 arctan (x/2f)

FOV ∝ 1 / f

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Focal length (f) and field of view (FOV) of a lens are inversely proportional.

From the equation we can say that,

A shorter focal length gives you a wide angle of view which allows more view to fit in the frame.

Hence,

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

Learn more about focal length here

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2 years ago
What is the energy of moving electrical charges
Simora [160]
The energy of moving electrical charges is Electrical energy



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3 years ago
At what point in the swing of the pendulum is the potential energy completely converted into Kinetic energy
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3 years ago
Read 2 more answers
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0
3 years ago
What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall
andrey2020 [161]

Answer:

D_{s} ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

     I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere               Is_{cm} = 2/5 M R²

Spherical shell   Ic_{cm} = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

    I = I_{cm} + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

     D = 2R

    Ic = Ic_{cm} + M D²

    Ic = 2/3 MR² + M (2R)²

    Ic = M R² (2/3 + 4)

    Ic = 14/3 M R²

The sphere

    Is =Is_{cm} + M [D_{s}²

    Is = Ic

    2/5 MR² + M D_{s}² = 14/3 MR²

    D_{s}² = R² (14/3 - 2/5)

    D_{s} = √ (R² (64/15)

    D_{s} = 2,066 R

3 0
3 years ago
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