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2 years ago

Which is the best example of tropism in plants?

Physics

1 answer:

seraphim [82]2 years ago

3 0

Answer:

Here are some examples of tropism in plants : Sunflowers are one of the best examples of positive phototropism , as they always grow facing the sun.

Explanation:

Although some experts consider that perhaps such a clear growth movement should not be considered tropism.

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A machine has a mechanical advantage of 5. What force should be applied to the machine to make it apply 3000 N to an object?

ziro4ka [17]

The answer to this very question is 3000n

6 0

3 years ago

Read 2 more answers

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.

romanna [79]

Answer:

(a): $\rm -5.627\times 10^3\ N.$

(b): $\rm 7.626\times 10^2\ N.$

Explanation:

Given:

Charge on one sphere, $\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.$

Charge on second sphere, $\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.$
Separation between the spheres, $\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.$

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}$

where,

k is called the Coulomb's constant, whose value is $\rm 9\times 10^9\ Nm^2/C^2.$

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.$

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, $\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.$

The new charges on both the spheres are equal and given by

$\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.$

The magnitude of the force that each sphere now experiences is given by

$\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.$

7 0

3 years ago

Which of the following is a true statement?

lukranit [14]

Answer:

LAST ONE

Explanation:

6 0

3 years ago

Read 2 more answers

9)A 64 kg parent and a 16 kg child meet at the center of an ice rink. They place their hands together and push. (A) Is the

Colt1911 [192]

Answer:

(A) The same

(B) More

Explanation:

The given parameters are;

The mass of the parent, m₁ = 64 kg

The mass of the child, m₂ = 16 kg

(A) By Newton's third law of motion, action and reaction are equal and opposite

Therefore, the action of the parent on the child is equal to the reaction of the child on the parent and vice versa

Therefore, the force experienced by the child is the same as the force experienced by the parent

(B) Newton's second law states that an objects acceleration is directly proportional to the applied force and inversely proportional to the mass of the object

Therefore, the parent and the child both experience the same force but the mass of the child is less than the mass of the parent and therefore, by Newton's second law, the acceleration of the child will be more than the acceleration of the parent for the same given force

Let F₁ represent the force experienced by the parent, let F₂ represent the force experienced by the child and let a₁ represent the magnitude of the parent's acceleration

By Newton's third law, we have;

F₁ = F₂

Force, F = Mass, m × Acceleration, a

We can write, F = m × a

Therefore;

F₁ = m₁ × a₁ and F₂ = m₂ × a₂

∴ F₁ = F₂ gives;

m₁ × a₁ = m₂ × a₂

a₁ = (m₂ × a₂)/m₁ = (16 × 2.5)/64 = 0.625

∴ The magnitude of the parent's acceleration = a₁ = 0.625 m/s²

5 0

3 years ago

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago