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Bad White [126]
2 years ago
5

A child blows a leaf straight up in the air from rest. The leaf accelerates at 1.0\,\dfrac{\text m}{\text s^2}1.0 s 2 m1, point,

0, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction upward.How much time does it take the leaf to displace 1.0\,\text m1.0m1, point, 0, space, m upwards?
Physics
2 answers:
pogonyaev2 years ago
8 0

Answer:

1.4

Explanation:

if its 1 m/s then just get the root of 2

Neko [114]2 years ago
3 0

Answer:

Time taken by the leaf to displace by 1.0 m distance is

t = \sqrt2 seconds

Explanation:

As we know that initial velocity of the leaf is given as

v_i = 0

now the acceleration upwards for the leaf is

a = 1 m/s^2

The displacement of leaf in upward direction is

d = 1 m

so now we have

d = v_i t + \frac{1}{2}at^2

1 = 0 + \frac{1}{2}(1) t^2

t = \sqrt2 seconds

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A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

\frac{1}{f} = (nglass - ni)(\frac{1}{R1} - \frac{1}{R2}).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

\frac{1}{f} = (1.5 - 1.3)(\frac{1}{10} - \frac{1}{15})

\frac{1}{f} = 0.2(\frac{1}{30})

f = \frac{30}{0.2} = 150

For air (nair = 1):

\frac{1}{f} = (1.5 - 1)(\frac{1}{10} - \frac{1}{15})

f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

5 0
2 years ago
Read 2 more answers
In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6
stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

                                      r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

                                      r₂= 7.5 cm

head upto water level is to rise = 60 + 5

                                          = 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q  

 π r₁² v₁ = π r₂² v₂  = 0.015

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{7.5^2}{5^2} v_2

 v_1= 2.25 v_2

 v_2 = \dfrac{0.015}{\pi r_2^2}

 v_2 = \dfrac{0.015}{\pi 0.075^2}

    v₂ = 0.848 m/s

    v₁ = 1.908 m/s

Applying Bernoulli's equation

 P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h

 P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65

 P_p= 635539.32 Pa

 P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

6 0
2 years ago
A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leave
balu736 [363]

Answer:

1.176 N

Explanation:

m = mass of the bottle = 0.30 kg

v_{o} = initial speed of the bottle = 2.8 m/s

v = final speed of the bottle = 0 m/s

d = stopping distance traveled = 1.0 m

f = magnitude of frictional force acting on bottle

Using work-change in kinetic energy theorem

- f d = (0.5) m (v^{2} - v_{o}^{2} )\\- f (1) = (0.5) (0.30) (0^{2} - 2.8^{2} )\\-f = - 1.176 \\f = 1.176 N

direction :

frictional force acts in opposite direction of motion.

8 0
2 years ago
The solar energy strikes the deck at the rate of 1400 W on every square metre.
Elena-2011 [213]

br[ o o o o o o o o o o o o o o o o o o o o opppppo o o o. oo o o o

8 0
1 year ago
A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo
Digiron [165]
Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
8 0
2 years ago
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