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murzikaleks [220]

3 years ago

13

What is the highest temperature ever recorded on earth

1 answer:

Llana [10]3 years ago

4 0

Therefore the world's record high temperature of 134.0°F (56.7°C) is held by Furnace Creek Ranch in Death Valley, California. That global high temperature was attained on July 10, 1913.

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Why is the unit of work is a derived unit<br>

masha68 [24]

Answer:

Since energy can be measured as work, we can write energy = force x distance. Thus SI derived unit of energy has the units of newtons x meter or kg m2/s2.

Explanation:

5 0

3 years ago

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Alli was in the park playing on the equipment. she noticed that on the highest slide she slides down

LUCKY_DIMON [66]

Answer:

Is this answer complete????

3 0

3 years ago

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A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following fun

snow_lady [41]

The correct answer would be 1.375 < t < 3 i hope this helps anyone

8 0

3 years ago

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A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a

Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial Velocity of Each object

$V_f$ = Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0

4 years ago

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}$

F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

$q_3=4.89\times 10^{-6}\ C$

Hence, this is the required solution.

3 0

3 years ago