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lord [1]
3 years ago
7

A pumpkin is thrown horizontally off of a building at a speed of 2.5 — and travels a horizontal distance of

Physics
1 answer:
Maurinko [17]3 years ago
3 0

Answer:

Explanation:

This is a missile throwing exercise, let's find the distance

         x = v₀ₓ / t

         t = v₀ₓ / x

let's calculate

         t = 2.5 / 12

         t = 0.2083 s

as time is a scalar this is the same value for descends to the ground

         y = v_{oy} t  - 1/2 g t²    

we calculate

         v_y = 0 - 9.8 0,2083  

         v_y = - 2.04 m / s

the negative sign indicates that the speed is down

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Uncle Harry weighs 750 N. What is his mass in kg?
Elenna [48]

Answer:

W = MG

750 = M * 10

M = 750/10

M = 75 kg

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3 years ago
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When is the universal theme of a story often revealed?
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It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>
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In what ways would organisms living in very deep water need to be specialized?
Nady [450]

Organisms living in great depths of water bodies like oceans and lakes need to be adapted for two (2) things especially; high water pressure and vision in darkness

The water column above from deep in the water can cause lots of hydrostatic pressure on the organisms’ cells. Also the fact that light cannot penetrate to great depth of water due to diffusion means the organisms must live in darkness.

Explanation:

It has been shown that cells from Piezophile bacteria have a high percentage of fatty acids in their membranes to prevent the cells from compacting solid from the high pressure.

Most of the organisms are also detritivores and use chemosynthesis for the autotrophs because light cannot reach these depths and hence photosynthesis is not possible.  Organisms with eye vision are adapted to high wavelength light that can at least reach greater depths before diffusing. Nonetheless natural selection would favour use of sight for most organisms in this benthic region.

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8 0
4 years ago
A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
3 years ago
2. A 50 kg diver is standing on the edge of a 15 m high cliff. What is his potential energy?
Lera25 [3.4K]

Answer:

3675 J

Explanation:

Gravitational Potential Energy = \frac{1}{2} × mass × g × height

( g is the gravitation field strength )

Mass = 50 kg

G = 9.8 N/kg ( this is always the same )

Height = 15 m

Gravitational Potential Energy = \frac{1}{2} × 50 ×9.8 × 15

= 3675 J

3 0
3 years ago
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