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Paladinen [302]
3 years ago
12

Q: A: The long distance water movement, from roots up to leaves, of the water transport in large plants depends on one unique pr

operty of water. Water's unique property of ______________ allows water molecules to stick together so they can be pulled upward by the force of transpiration. A) adhesion B) cohesion C) indention D) retention
Physics
2 answers:
Gnom [1K]3 years ago
5 0
A. Cohesion
Because adhesion is the property of attraction to a different substance but cohesion is the tendency of water molecules to attract one another
ICE Princess25 [194]3 years ago
4 0
Answer: B

Cohesion makes water stick together, which is vital in transporting nutrients from tree roots to other parts of the plant.
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Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

4 0
4 years ago
Two blocks with the same weight but different dimensions are floating in water at different levels.
fredd [130]

Answer:

(ii) less than

Explanation:

The blocks has the same mass. So, the buoyant force depends only on the volume of the displaced fluid. Since block A has a smaller volume than block B, it displaces a smaller amount of fluid. Therefore the bouyant force exerted by water is less in block A than in block B.

6 0
4 years ago
1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Cal
disa [49]

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given:

- The radius of earth r = 6.3781×10^6 m

- The angular speed of earth w = 7.27*10^-5 rad/s

- The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

- The mass of earth m = 5.972 × 10^24 kg

- The inertia of sphere I = 2/5 * m* r^2

Solution:

- The angular acceleration of the earth from rest to w is given by α:

                                α = w / t

                                α = (7.27*10^-5) / (1.57784760 * 10^8)

                                α = 4.60754*10^-13 rad/s^2

- The required torque τ is given by:

                                τ = I*α

                                τ = 2/5 * m* r^2 * α

                           τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

                            τ = 4.47746 * 10^25 N-m

- The power required P to turn the earth to the speed w is:

                           P = τ*w

                           P = (4.47746 * 10^25)*(7.27*10^-5)

                           P = 3.25511*10^21 W

- The energy E required is :

                           E = P / t

                           E = (3.25511*10^21) / (1.57784760 * 10^8)

                           E = 2.06301 * 10^13 J

4 0
3 years ago
A volleyball is served at a speed of 8 / at an angle 35° above the horizontal. What is the speed of the ball when received by th
bezimeni [28]
The speed of the ball is how hard u hit it
7 0
3 years ago
I need help with this one <br><br> Coulombs law <br><br> 100 points and brainliest
Ganezh [65]

Answer:

my dad is a physics professor so when he comes home i will tell him to answer the question for you

3 0
3 years ago
Read 2 more answers
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