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3 years ago

How Is budding different from fertilization

Physics

1 answer:

tiny-mole [99]3 years ago

5 0

<span>Artificial fertilizers are made chemically. They emphasize three main</span>

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PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!! 80 POINTS!!!!!!!!!!!!

Alona [7]

Answer: clay

Explanation:

Florida is mainly influences by Spanish culture, so building with clay and wood would make sense, knowing that that is how pueblas are built.

8 0

3 years ago

Read 2 more answers

A small toy car travels 2.0 meters in 120 seconds. what is the speed of the car?

mamaluj [8]

Answer: $\dfrac{1}{60} ms^{-1}$

Explanation:

Speed of an object is defined as the ratio of the distance covered by the object to the time taken to cover that distance.

Let $s$ be the speed of the object.

Let $D$ be the distance travelled by the object.

Let $t$ be the time taken by the object.

So, $s=\frac{D}{t}$

s= $\frac{2}{120}=\frac{1}{60}ms^{-1}$

So,the speed of the car is $\frac{1}{60}ms^{-1}$

4 0

3 years ago

An object of mass 2 kg starts from rest and is allowed to slide down a

WARRIOR [948]

Answer: B

Explanation:

Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.

The parameters given from the question are:

Mass M = 2kg

Height h = 20m

Let g = 9.8m/s^2

At the bottom of the incline plane, the object will experience maximum kinetic energy.

From conservative of energy, maximum K.K.E = maximum P.E

Maximum P.E = mgh

Maximum P.E = 2 × 9.8 × 20 = 392 J

But

K.E = 1/2mv^2

Substitute the values of energy and mass into the formula

392 = 1/2 × 2 × V^2

V^2 = 392

V = sqrt( 392 )

V = 19.8 m/s

V = 20 m/s approximately

7 0

3 years ago

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

$sin\theta - \mu_k cos\theta = 1$

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

5 0

3 years ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago