To find the angle, in radians, that the ladder makes with the ground, we can use trigonometry. The length of the ladder can be found using the Pythagorean theorem, and the angle can be calculated as the arcsine of the ratio of the height of the building to the length of the ladder. The angle is approximately 0.6435 radians.
Explanation:To find the angle, in radians, that the ladder makes with the ground, we can use trigonometry. The ladder, the ground, and the building form a right triangle. The ratio of the opposite side (the height of the building) to the hypotenuse (the length of the ladder) is equal to the sine of the angle. Using this information, we can calculate the angle in radians.
First, we need to find the length of the ladder using the Pythagorean theorem: l^2 = 20^2 + 15^2 = 625. Taking the square root of both sides, we find that the length of the ladder is 25 feet.
The sine of the angle can be calculated as the ratio of the height of the building to the length of the ladder: sin(angle) = 15/25 = 0.6. Taking the arcsine (inverse sine) of 0.6, we find that the angle in radians is 0.6435 (rounded to two decimal places).
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Two football players are pushing a 60 kg blocking sled across the field at a constant speed of 2.0 m/s. The coefficient of kinetic friction between the grass and the sled is 0.30. Once they stop pushing, how far will the sled slide before coming to rest?
Answer:
The sled slides d=0.155 meters before rest.
Explanation:
m= 60 kg
V= 2 m/s
μ= 0.3
g= 9.8 m/s²
W= m * g
W= 588 N
Fr= μ* W
Fr= 176.4 N
∑F = m * a
a= (W+Fr)/m
a= 12.74m/s²
t= V/a
t= 0.156 s
d= V*t - a*t²/2
d= 0.155 m
An electron with kinetic energy 2.9 keV moving along the positive direction of an x axis enters a region in which a uniform electric field of magnitude 7.5 kV/m is in the negative direction of the y axis. A uniform magnetic field is to be set up to keep the electron moving along the xaxis, and the direction of the field is to be chosen to minimize the required magnitude of the field. What is the magnitude of the magnetic field in mT?
The electric force on the electron is opposite in direction to the electric field E. E points in the -y direction, so the electric force will point in the +y direction. The magnitude of the electric force is given by:
F = Eq
F = electric force, E = electric field strength, q = electron charge
We need to set up a magnetic field such that the magnetic force on the electron balances out the electric force. Since the electric force points in the +y direction, we need the magnetic force to point in the -y direction. Using the reversed right hand rule, the magnetic field must point in the -z direction for this to happen. Since the direction is perpendicular to the +x direction of the electron's velocity, the magnetic force is given by:
F = qvB
F = magnetic force, q = charge, v = velocity, B = magnetic field strength
The electric force must equal the magnetic force.
Eq = qvB
Do some algebra to isolate B:
E = vB
B = E/v
Let's solve for the electron's velocity. Its kinetic energy is given by:
KE = 0.5mv²
KE = kinetic energy, m = mass, v = velocity
Given values:
KE = 2.9keV = 4.6×10⁻¹⁶J
m = 9.1×10⁻³¹kg
Plug in and solve for v:
4.6×10⁻¹⁶ = 0.5(9.1×10⁻³¹)v²
v = 3.2×10⁷m/s
B = E/v
Given values:
E = 7500V/m
v = 3.2×10⁷m/s
Plug in and solve for B:
B = 7500/3.2×10⁷
B = 0.00023T
B = 0.23mT
To minimize the required magnitude of the magnetic field and keep the electron moving along the x-axis, the direction of the magnetic field should be chosen to cancel out the force due to the electric field. The magnitude of the magnetic field needed is 2.59 mT.
Explanation:To keep the electron moving along the x-axis and minimize the required magnitude of the magnetic field, the force on the electron due to the magnetic field should cancel out the force on the electron due to the electric field.
The force on the electron due to the electric field is given by: