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2 years ago

9

Horseshoe crabs, or Limulus polyphemus, are very common along the east coast of the US. Fossils of horseshoe crabs have revealed

that they have changed very little in the past 300 million years. Which statement best explains why this is the case?
A.Horseshoe crabs have a low reproductive rate.

B.Horseshoe crabs need to change only if humans are present in their environment.

C.The horseshoe crab is well adapted to its environment. and its environment has not changed much in the last 300 million years.

D.Horseshoe crabs have a high mutation rate.

2 answers:

liubo4ka [24]2 years ago

5 0

The answer to your question is C

AnnyKZ [126]2 years ago

4 0

The answer to this question is C

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What is the net charge of a metal ball if there are 21,749 extra electrons in it?

pickupchik [31]

Answer:

$Q=3.47\times 10^{-15}\ C$

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is $q=1.6\times 10^{-19}\ C$

To find the net charge if there are n number of extra electrons is :

Q = n × q

$Q=21749\times 1.6\times 10^{-19}\ C$

$Q=3.47\times 10^{-15}\ C$

So, the net charge on the metal ball is $3.47\times 10^{-15}\ C$ . Hence, this is the required solution.

6 0

3 years ago

A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1

SCORPION-xisa [38]

Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

5 0

2 years ago

A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$

$v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)$

$v_f^2 = 567.9$

$v_f = 23.8 m/s$

Part c)

Relative speed of two balls is given as

$v_{12} = v_1 - v_2$

$v_{12} = (13.9) - (-13.9) = 27.8 m/s$

now the distance between two balls in 0.8 s is given as

$d = v_{12} t$

$d = 27.8 \times 0.8$

$d = 22.24 m$

7 0

3 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago