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mafiozo [28]
2 years ago
8

What’s 2.3 miles into kilometers

Physics
1 answer:
True [87]2 years ago
8 0
3.701 kilometers hope that helps
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How many types of quarks are there?<br> a. 2<br> b. 4<br> c. 6<br> d. 8
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There are 6 types of quarks.

Answer: C. 6
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3 years ago
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A 7 kg ball is moving at a constant speed of 5 m/s. A force of 300 N is applied to the ball for 4 s. The new speed of the ball i
olasank [31]

Answer:

The new speed of the ball is 176.43 m/s

Explanation:

Given;

mass of the ball, m = 7 kg

initial speed of the ball, u = 5 m/s

applied force, F = 300 N

time of force action on the ball, t = 4 s

Apply Newton's second law of motion;

F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v =  \frac{Ft}{m} + u

where;

v is new speed of the ball

v =  \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s

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3 years ago
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
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Fertilizers usually contain two elements from Group 5A, which are
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