A 7kg block slides down a 30 degree slope at a constant velocity of 2 m/s. How big is the force of friction acting on the block?
1 answer:
The force of friction is <u>34.3 N.</u>
A block of mass m slides down a plane inclined at an angle θ to the horizontal with a constant velocity. According to Newton's first law of motion, every body continues in its state of rest or a state of uniform motion in a straight line, unless acted upon, by an external unbalanced force. This means that when balanced forces act on a body, the body moves with a constant velocity.
The free body diagram of the sliding block is shown in the attached diagram. Resolve the weight mg of the block into two components mg sinθ along the direction of the plane and mg cosθ perpendicular to the plane . The force of friction F acts upwards along the plane and the normal reaction acts perpendicular to the plane.
Since the block moves down with a constant velocity, the downward force mg sinθ must be equal to the upward frictional force.
Substitute 7 kg for m, 9.8 m/s² for g and 30° for θ.
The force of friction is <u>34.3 N</u> up the plane.
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Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
Next, you use the equation (2) and solve for λ:
C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
hence, the radius of the atom in its 5-th state is 59.5 anstrongs