Answer:
The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?
f= Force (of gravity)=500N
g=acceleration of gravity=1.6m/s^2
m=mass=312kg
m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg
his mass is 312kg
The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
To find the answer, we need to know about the magnetic field inside the solenoid.
<h3>What's the expression of magnetic field inside a solenoid?</h3>
- Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
- n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
- Here, n = 290/32cm or 290/0.32 = 906
I= 0.3 A
- So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.
Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
Learn more about the magnetic field inside the solenoid here:
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Answer:
The speed of transverse waves in this string is 519.61 m/s.
Explanation:
Given that,
Mass per unit length = 5.00 g/m
Tension = 1350 N
We need to calculate the speed of transverse waves in this string
Using formula of speed of the transverse waves

Where,
= mass per unit length
T = tension
Put the value into the formula


Hence, The speed of transverse waves in this string is 519.61 m/s.
I think frequency it sounds like the correct answer but I am not completely sure if I am correct
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)