All categories

3 years ago

Why have fungi evolved mechanisms for dispersal of their spores

1 answer:

3 0

Not all fungi did that. At one time, there were fungi that dispersed
their spores and fungi that did not disperse their spores.

The fungi that did not disperse their spores produced very few
new fungi, and they aren't around any more.

You might be interested in

In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for

Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0

3 years ago

The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o

blsea [12.9K]

Answer:

$F_f=840N$

Explanation:

From the question we are told that

Weight of fireman $W_f= 87.5kg$

Pole distance $D=4.10m$

Final speed is $V_f 1.75m/s$

Generally the equation for velocity is mathematically represented as

$v^2 = v_0^2 + 2 a d$

Therefore Acceleration a

$a'= v^2 / 2 d$

$a'= 0.21m/s^2$

Generally the equation for Frictional force $F_f$ is mathematically given as

$F_f=m*a$

$F_f=m*(g-0.21)$

$F_f=87.5*(9.81-0.21)$

Therefore

$F_f=840N$

6 0

3 years ago

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c

Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction $\mu =0.8$

Angle of inclination is equal to $\Theta =tan^{-1}\mu$

$\Theta =tan^{-1}0.8=38.65^{\circ}$

$tan{38.65^{\circ}}=0.8$

Radius is given r = 28 m

Acceleration due to gravity $g=9.8m/sec^2$

We know that $tan\Theta =\frac{v^2}{rg}$

$0.8=\frac{v^2}{28\times 9.8}$

$v^2=219.52$

$v=14.816m/sec$

So before start of slide velocity will be 14.81 m/sec

3 0

3 years ago

The Venn diagram compares protons with neutrons. Which shared property belongs in the region marked "B"?

telo118 [61]

is there any choices?

4 0

3 years ago

Read 2 more answers

ly charged particles are held 24 x 103m apart and then released from rest. The initial acceleration of the first particle is obs

inessss [21]

Answer:

Part a)

$m_2 = 4.9 \times 10^7 kg$

Part b)

$q_1 = q_2 = 5312.6 C$

Explanation:

Part a)

As we know that both charge particles will exert equal and opposite force on each other

so here the force on both the charges will be equal in magnitude

so we will have

$F = m_1a_1 = m_2a_2$

here we have

$6.3 \times 10^7(7) = m_2(9)$

now we have

$m_2 = 4.9 \times 10^7 kg$

Part b)

Now for the force between two charges we can say

$F = \frac{kq_1q_2}{r^2}$

now we have

$(6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}$

now we have

$q_1 = q_2 = 5312.6 C$

3 0

3 years ago