A typical land snail's speed is 12.2 meters per hour. How many miles will the snail travel in one day(24hrs)?

Question 3 of 10

AlexFokin [52]

The answer is Jupiter

6 0

2 years ago

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{WILL GIVE (Brain lies t) TO FULL ANSWERS}

kap26 [50]

Answer:

1: surface temperature

2: red giant

3: The brightest stars are called supergiants. Star clusters are rich in stars just off the main sequence called red giants. Main sequence stars are called dwarfs.

4: A white dwarf is very dense

5: red giant

plz mark me as brainliest :)

4 0

2 years ago

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One side of the moon always faces Earth because the time it takes the moon to spin on its axis is blank the time it takes the mo

Kamila [148]

The word to fill in the blank is "equal". Because the time taken to rotate (spin on its axis) is equal to the time of revolution (going around the earth), this means that both have the same rate of angular rotation. So for every bit that the moon goes around its orbit around earth, the moon itself rotates accordingly to present the exact same side to earth.

6 0

3 years ago

Calculate the TOTAL mechanical energy of pendulum is it swings from his highest point to its lowest point. Pendulum mass is 4

Lerok [7]

Answer:

its should be 2.0 and 4.5 on it

3 0

3 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

2 years ago