Answer:
The magnitude will be "353.5 N". A further solution is given below.
Explanation:
The given values is:
F = 500 N
According to the question,
In ΔABC,
⇒ 
⇒ 
then,
⇒ 
⇒ 
Now,
The corresponding angle will be:
⇒ 
⇒ 
⇒ 
Aspect of F across the AC arm will be:
= 
On putting the values of F, we get
= 
= 
Component F along the AC (in magnitude) will be:
= 
= 
= 
Answer:
its In most solids, conduction takes place as particles vibrate in place.
Explanation:
i did it in a test
The emf induced = B*l*v where B is the flux density, l the length of the conductor and v the velocity of the conductor. In the given case B = 0.035 N/amp.meter, l = 0.86 and v = 6 m/sec
emf = 0.035*0.86*6 = 0.1806 v ≈ 0.18 v
choice: D
Answer:
The travel would take 6.7 years.
Explanation:
The equation for an object moving in a straight line with acceleration is:
x = x0 + v0 t + 1/2a*t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
In a movement with constant speed, a = 0 and the equation for the position will be:
x = x0 + v t
where v = velocity
Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74 m/s²:
Seconds in half a year:
1/2 year = 1.58 x 10⁷ s
x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)² = 1.59 x 10¹⁵ m
Now let´s see how much time it takes the travel to the nearest star after this half year.
The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²
v = v0 + a t
Since the spacecraft starts from rest, v0 = 0
v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s
Using the equation for position:
x = x0 + v t
4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t
(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t
t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.
The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.
Answer:
Explained
Explanation:
1.Each of the spring scale will read 10N,considering acceleration due to gravity as 10 m/s^2
2.Each of the spring scale will read 10N because each string exerts a force of 10 N to counterbalance the force of 1 kg mass attached to it. This means the tension on the both side of the string is 10 N. So the scale will read 10 N. Also as spring balances are attached in series and kept on table so both spring balances will read same readings.
