All categories

3 years ago

When I say that in Australia there is less rain and is cooler in average, am I talking about weather or climate?

Physics

1 answer:

svetoff [14.1K]3 years ago

8 0

Answer: climate

Explanation:

You might be interested in

A horse slows from a velocity of 9.5 m/s to 5.5 m/s over a distance of 32 m. Find time.

lesya [120]

Answer:

Explanation:

Finding time= t=change in velocity or distance/acceleration

so 32/9.5-5.5

32/4

6 0

4 years ago

Calculate the potential difference between points x and y

aliya0001 [1]

Answer:

4.275v

<u><em>Thank you </em></u>

8 0

3 years ago

Read 2 more answers

(PLEASE HELP! 30 POINTS!) What are two similarities and two differences between the Strong Nuclear Force and Weak Nuclear Force?

pickupchik [31]

Answer:

Difference: The weak force tears an atomic nucleus apart.The strong force is what keeps the protons and neutrons in a nucleus held together, while the weak force is what makes radioactive particles decay.Or even simpler: The strong force keeps stuff together while the weak force helps stuff split up.

Similarities: Weak force allows quarks to turn into other quarks, which can turn a neutron into a proton and break up the nucleus. ... One big difference is that the particles/fields that are responsible for carrying the strong force are massless, whereas for the weak force they are quite massive (nearly 100 times the proton mass).

5 0

4 years ago

Calculate the total resistance for a 650ohm , a 350 ohm , and a 1000 ohm resistor connected in series

Mekhanik [1.2K]

Answer:

2000 ohms

Explanation:

Resisters in series just add.

Rt = R1 + R2 + R3

R1 = 650 ohm

R2 = 350 ohm

R3 = 1000 ohm

Rt = 650 + 350 + 1000

Rt = 2000 ohms.

5 0

3 years ago

Read 2 more answers

The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are

r-ruslan [8.4K]

Answer with explanation:

Part a)

$v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s$

Part b)

$a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}$

8 0

3 years ago