Answer:
1. a
2. a [im iffy on this but 95% positive its this]
3. b [walking is a form of aerobics, so i would say b]
Explanation:
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: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
When using chlorine bleach, the proper amount for one gallon of water is 8 ounces, or one cup. Now, since there's 4 quarts in a gallon,divide the 8 ounces by 4. You get 2. So you should put 2 ounces (or 1/4 cup) of chlorine bleach in a quart of water. Hope this helps!
Answer:
speed of star is 1.37 ×
m/s
it is approaching to earth because wavelength of star is decreasing than rest
if emit same wavelength it does not move anywhere
it will remain steady condition with respect earth
Explanation:
given data
wavelength = 656.5 nm
rest wavelength = 656.8 nm
to find out
the speed of the star , is it approaching
solution
we know here equation that is
wavelength shift / wavelength at rest = velocity of source / speed of light
so put all value and find
velocity of source = 3 ×
× 3 ×
/ 656.8 × 
velocity of source star = 1.37 ×
m/s
and
it is approaching to earth because wavelength of star is decreasing than rest
and
if emit same wavelength it does not move anywhere
it will remain steady condition with respect earth