A string is stretched by two equal and opposite forces F newton each then the tension in sting is?

A cylinder of radius R and height H is floating upright in

emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

$F = mg$

so we have

$\Delta P \pi R^2 = mg$

so we have

$\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g$

so we have

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

6 0

2 years ago

1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe

sdas [7]

Hypothesis: The water will squirt far.
Indep.V.: Height of hole.
Depend.V.: Range of squirt
Constant: Everything that isnt the independant var. such as filled liquid.
Control: None,I believe.
Number of groups: 4
Trials per group: 4

7 0

2 years ago

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Suppose you see two main-sequence stars of the exact same spectral type. Star 1 is dimmer in apparent brightness than Star 2 by

katrin2010 [14]

Options:

A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.

B. Star 1 is 100 times more distant than Star 2.

C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.

D. Star 1 is 10 times more distant than Star 2.

E. Star 1 is 100 times nearer than Star 2.

Answer:

D. Star 1 is 10 times more distant than star 2

Explanation:

For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.

Luminosity, L = 1/r²

Where r is the distance of the star to the earth

Since star 1 is dimmer in brightness than star 2 by a factor of 100,

L₁/L₂ = 1/100

i.e. L₁ = 1, L₂=100

L₁ = 1/r₁² ............(1)

1 = 1/r₁²

L₂ = 1/r₂²

100 = 1/r₂² .........(2)

divide equation (2) by equation (1)

100/1 = ( 1/r₂² )/ (1/r₁²)

100 = (r₁/r₂)²

r₁/r₂ = √100

r₁/r₂ = 10

r₁ = 10r₂

3 0

1 year ago

A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops

Anit [1.1K]

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

4 0

1 year ago

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A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m

tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/ (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0

1 year ago