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3 years ago

13

What side of the worm faces up when placed in the tray when dissection?

1 answer:

Lelechka [254]3 years ago

8 0

<span>The side of the worm that faces up when placed in the tray for dissection is the smoother side.

</span>You can observe the organs of these tiny creatures by dissecting a preserved earthworm<span>. In dissecting a worm, </span><span>lay the worm on your </span>dissecting tray<span> <span>with its dorsal side facing up.</span></span>

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A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 meters from the wall and the sound of h

ValentinkaMS [17]

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

8 0

2 years ago

a baseball player is dashing toward home plate with a speed of 200 feet per hour. When he decides to hit the dirt. he slides for

Serggg [28]

To solve this problem we will use the kinematic formula for the final velocity.

$V_f_x = V_0_x + a_xt$

The final speed is 0 at the moment the player stops.

The time until it stops is 1.3 s

The initial speed is 200 feet / s Note (check the speed units in the problem statement, 200ft / s is very much and 200ft / h is very small)

Then, we clear the formula.

$a_x = \frac{(Vfx-V0x)}{t}\\ a_x = \frac{(0-200)}{1.3}\\ a_x = -153.5 ft / s ^ 2$

Because the player is slowing down, the acceleration goes in the opposite direction to the player's movement, and that is why it is negative.

To answer part b) we use the following formula.

$Vf ^ 2 = Vo ^ 2 + 2a * (x_2 - x_1)\\\\ (x_2 - x_1)= \frac{(V_f ^ 2-V_0 ^ 2)}{2a}\\\\ (x_2 - x_1)= \frac{(0-200 ^ 2)}{- 2 * 153.5}\\\\ (x_2 - x_1)= 130.29 feet$

8 0

2 years ago

Kipish [7]

Answer:

The intensity of the electric field is

$|E|=10654.37 \:N/C$

Explanation:

The electric field equation is given by:

$|E|=k\frac{q}{d^{2}}$

Where:

k is the Coulomb constant
q is the charge at 0.4100 m from the balloon
d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

$|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}$

$|E|=10654.37 \:N/C$

I hope it helps you!

5 0

2 years ago

If you wanted to learn more about what it was like to work in a family of scientists, whose life would you research?

Schach [20]

Answer:

Stephen hawking if his family were scientists

6 0

3 years ago

Read 2 more answers

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

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

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago