A uniform ladder is 10 m long and weighs 400 n. it rests with its upper end against a frictionless vertical wall. its lower end
rests on the ground and is prevented from slipping by a peg driven into the ground. the ladder makes a 30◦ angle with the horizontal. the magnitude of the force exerted on the peg by the ladder is:
1 answer:
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Answer:
0.010 m
Explanation:
So the equation for a pendulum period is: where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:
Evaluate the multiplication in front
Divide both sides by 6.28
Square both sides
Multiply both sides by m/s^2 (the s^2 will cancel out)
Now now let's find the length when it's two seconds
Divide both sides by 6.28
Square both sides
Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)
So to find the difference you simply subtract
0.984 - 0.994 = 0.010 m
Explanation:
Here is the answer,
Lets be BF
Answer:
The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction
Explanation:
In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:
(1)
v: speed of the proton = 9.9*10^5 m/s
q: charge of the proton = 1.6*10^-19C
B: magnetic field = ?
FB: magnetic force on the proton = 1.6*10^-13N
When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:
^j X (-^i) = -(-^k)=^k
To obtain the magnitude of the magnetic field you use:
The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction