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tatuchka [14]
3 years ago
13

What part of the plant transports water from the roots to the leaves?

Physics
1 answer:
shusha [124]3 years ago
7 0
<span><span>1.       </span>Question : What part of the plant that transport water from its root to its leaves.
The answer is the stem. The stem connects the plant’s root and leaves, it is responsible for transporting water that is absorbed by the root going to the leaves, it is also responsible for transporting nutrients absorbed with the leaves coming from the sun or any nutrients going to the roots to make is grow faster and healthy. Stem is the one that connects leaves and root and the one the does the transporting job.</span>



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A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the
marysya [2.9K]

Answer:

Explanation:

1 ) the volume of helium in the balloon when fully inflated

= 4/3 π R³

= 4/3 x 3.14 x 5.3³

= 623.3 m³

2 ) force of gravity on the entire system ,but with no people

= (mass of the empty balloon and basket + mass of helium )x g

=( 129 +  623.3 x.18 )x 9.8

= 2363.7 N

3 )  magnitude of the buoyant force on the entire system (but with no people )

( Volume of empty balloon + basket + volume of helium ) density of air x g

( 0.057 + 623.3 ) x 1.23 x 9.8

= 7513.94 N

4 )  magnitude of the force of gravity on each person

Mass of each person x g

= 71 x 9.8

= 695.8 N

6 0
3 years ago
When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that yo
xz_007 [3.2K]

Answer: 0.1 m, 0.0583 m

Explanation:

We are given that:

Frequency for throat= 800 Hz

Frequency for mouth= 1500 Hz

Sound speed= 350 m/s

We have to find the corresponding lengths.

We have

f= \frac{v}{4L}

or L=\frac{v}{4f}

For the throat= L= \frac{350}{4*800\\} = 0.1 m

For the mouth= L= \frac{350}{4*1500} = 0.0583 m

3 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you.Your reaction time
shtirl [24]

Consider the motion of the car before brakes are applied:

v₀ = maximum initial velocity of the car before the brakes are applied

t = reaction time = 0.50 s

x₀ = distance traveled by the car before brakes are applied

since car moves at constant speed before brakes are applied

Using the equation

x₀ = v₀ t

x₀ = v₀  (0.50)


Consider the motion after brakes are applied :

v₀ = initial velocity of the car before the brakes are applied

a = acceleration = - 10 m/s²

v = final velocity of the car after it comes to stop = 0 m/s

x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀  (0.50)

Using the equation

v² = v²₀ + 2 a x

inserting the values

0² = v²₀ + 2 (- 10) (38 - v₀  (0.50))

v²₀ = 20 (38 - v₀  (0.50))

v₀ = 23 m/s



3 0
2 years ago
Two uncharged, conducting spheres are separated by a distance d. When charge −Q is moved from sphere A to sphere B, the Coulomb
rosijanka [135]

Answer: a) the force will be repulsive

b) the ratio of the new force to the old force will be 2

c) O

Explanation:

a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.

b) initial force will be -q(-Q)/d2

Adding extra charge -Q will cause change on B to become -2Q

The new force will be - 2Q(-q)/d2

Dividing new force by old force will give 2

C) if B is neutralized, the net charge becomes 0 and there will be no force on it.

3 0
2 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
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