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Oksi-84 [34.3K]
2 years ago
5

Multiply the following and write your answer using significant figures. What is 25 x 15?

Physics
1 answer:
Troyanec [42]2 years ago
3 0
25x15 is 375 cndnmekcivjfndn(sorry it said I needed 20 characters to comment)
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A 900-kg giraffe runs across a field at a rate of 50 km/hr. What is the magnitude of its momentum? 18 km/hr
Yuri [45]
The correct answer is the one with 45000 kg*km/hr.
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What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch?
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A)We know the formula of the angular speed is                 ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is          ω= 2π / (60s)             = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is             T = 1 hr                = 3600sThen the angular speed of the minute hand is         ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is          T = 12hrs             = (12)(3600)sThen the angular speed of the hour hand is         ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
8 0
3 years ago
Can someone please answer 5-7
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What is the evidence proving that we're in the 6th mass extinction?
Sveta_85 [38]

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8 0
2 years ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
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