How many degrees are in each quadrant <br> A: 90°<br> B: 30°<br> C: 180°<br> D: 360°

Which one is it?????

antoniya [11.8K]

Answer:ummm

Explanation:

masss

5 0

2 years ago

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a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration

slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0

2 years ago

g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

AnnyKZ [126]

Answer:

Time, $t=1.07\times 10^{-23}\ s$

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel $3.2\times 10^{-15}\ m$ before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

$t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s$

So, the time interval is $1.07\times 10^{-23}\ s$

5 0

3 years ago

How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se

Montano1993 [528]

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats

3 0

2 years ago

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An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

2 years ago

How many degrees are in each quadrant A: 90° B: 30° C: 180° D: 360°