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2 years ago

How would a continents climate change if it drifted closer to the equator?

1 answer:

8 0

The equator is the closest part of out planet that is closest to the sun, which means that the continents closest to the equator will in turn be warmer.

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Precipitation is greater over what and less over what?<br><br> A. Land, sea<br><br> B. Sea, land

Ksju [112]

I would say letter A to this because we need rain water for crops plants. Also we need it so that we can drink it. Its less great over sea because as it merge with the salt in the ocean it becomes less drinkable and less usable.

3 0

3 years ago

A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c

tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

$B=\dfrac{\mu_{0}NI}{l}$

Where, N = Number of turns

I = current

l = length

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}$

$B=0.005026=5.026\times10^{-3}\ T$

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

4 0

3 years ago

Stick a fork into each end of the root vegetable as shown. The forks should be on the same side of the vegetable, and

zhuklara [117]

Answer:

Yes

Explanation:

There is a position that works better than this and that is switching the sides of the forks.

7 0

2 years ago

When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s

d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(2)^2 = 0.5 J$

2) when speed is 3 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(3)^2 = 1.125 J$

3) when speed is 4 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(4)^2 = 2 J$

4) when speed is 5 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(5)^2 = 3.125 J$

5) when speed is 6 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(6)^2 = 4.5 J$

3 0

3 years ago

Read 2 more answers

A car battery has a rating of 250 ampere-hours. This rating is one indication of the total charge that the battery can provide t

Serga [27]

Answer:

Total charge provided by the battery could be 900000 C.

Maximum current provided by the battery for 37 minutes could be 405.405 A

Explanation:

Rating= 250 A-h

a. Total charge:

$Rating=Q/t\\Q=Rating.t\\$

Suppose t=1h

$Q=250 A(1h)\\Q=250 A(3600 sec)\\Q= 900000 A.sec$

We konw that $Ampere=\frac{Coulomb}{sec}$ , replacing:

$Q=900000(\frac{Coulomb}{sec})(sec)\\Q=900000 Coulomb$

Total charge provided by the battery could be 900000 C.

b. Maximum current for 37 minutes

$I=\frac{Q}{t} \\I=\frac{900000 C}{37*60 sec}\\I=405.405 A$

Maximum current provided by the battery for 37 minutes could be 405.405 A

4 0

3 years ago