Answer:
The hiker followed a road heading north for 2 miles in 30 minutes.
Explanation:
In order to describe the motion of an object, distance covered and time taken must be required. The total path covered by an object is called the distance travelled.
The hiker followed a road heading north for 2 miles in 30 minutes. This describes the motion of hiker. The motion shows how fast the hiker is moving.
Distance, d = 2 miles = 3218.6 m
times, t = 30 minutes = 1800 seconds
So, we can say that the hiker is moving with a speed of 1.78 m/s in north direction.
Hence, this is the required solution.
Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
Answer:
Velocity, v = 0.239 m/s
Explanation:
Given that,
The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m
The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.
For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.
Frequency is number of cycles per unit time.
Now we can find the velocity of the wave.
Velocity = frequency × wavelength
v = 0.574 × 0.418
v = 0.239 m/s
So, the velocity of the wave is 0.239 m/s.
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
