Answer:
maximum speed 56 km/h
Explanation:
To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose
sin 10 = Nx / N
cos 10 = Ny / N
Ny = N cos 10
Nx = N sin 10
Let's develop Newton's equations on each axis
X axis
We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve
Nx + fr = m a
a = v2 / r
fr = mu N
N sin10 + mu N = m v² / r
N (sin10 + mu) = m v² / r
Y Axis
Ny -W = 0
N cos 10 = mg
Let's solve these two equations,
(mg / cos 10) (sin 10 + mu) = m v² / r
g (tan 10 + μ / cos 10) = v² / r
v² = r g (tan 10 + μ / cos 10)
They ask us for the maximum speed
v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)
v² = 294 (0.8364)
v = √(245.9)
v = 15.68 m / s
Let's reduce this to km / h
v = 15.68 m / s (1 km / 1000m) (3600s / 1h)
v = 56.45 km / h
This is the maximum speed so you don't skid
"Apparent magnitude" means how bright a star looks to
a person on Earth.
-- The star that appears brightest is the one with the
lowest-number apparent magnitude . . . Star-C, at -4 .
-- All of them are visible from Earth, but may require some 'help'.
The dimmest stars visible with good human eyes under dark,
non-polluted skies are those with apparent magnitude around 6.
Stars B and C would be visible to the unaided eye, but Star-A
would require binoculars.
Around here, a few miles outside of the Chicago city limits, we're
lucky to see Magnitude-4 without binoculars.
-- It's not possible to determine which star has the highest luminosity.
The apparent magnitude depends on the star's distance from Earth
as well as its luminosity.
A flashlight 3 feet from your face appears much brighter than any
star, although any star is more luminous than the flashlight.
Distance from you has a lot to do with it.
_____________________________________________
"Absolute magnitude" means how bright each star would appear
to a person on Earth if all stars were at the same distance from us.
(The distance happens to be 32.6 light years.) It only depends on
the star's real luminosity, not on its distance.
-- It's not possible to determine which star appears brightest.
Star-C (absolute -7) would appear brightest if all stars were
equal distances from us. But a flashlight ... which has a huge-
number absolute magnitude because we couldn't see at all from
32.6 light years away ... can appear very bright from 3 feet in
front of your face.
-- They're all visible from Earth, but a star with absolute magnitude
greater than 6 would need binoculars (or better) to be visible.
-- Yes, if you know a star's absolute magnitude, then you know its
luminosity. The lowest-number absolute magnitudes are the ones
that would appear brightest if all stars were the same distance from
us, so they're the stars with the greatest luminosity. From this group,
that's Star-C.
Answer: There are many possible elements, and they are all in the same vertical column as bromine.
Explanation:
In a periodic table, the elements are arranged according to the atomic number. The elements arranged in the same vertical column (known as groups) have same valence configuration and therefore have same chemical properties. Hence, there would be more possible elements having same chemical properties in the same vertical column (group) as Bromine.
Answer:
mass of the planet X = 5.6 × 10²³ kg.
Explanation:
According to Newtons law of universal gravitation,
F = GM₁M₂/r²
Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X
making M₂ The subject of the equation above,
M₂ = Fr²/GM₁ .......................... equation 2
Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴ × 1000 m
r = 1.08 × 10⁷ m, G = 6.67 × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg
Substituting this values in equation 2,
M₂ = 24.13(1.08 × 10⁷ )²/75( 6.67 × 10 ⁻¹¹)
M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹
M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)
M₂ = 0.056 × 10²⁵
M₂ = 5.6 × 10²³ kg.
Therefore mass of the planet X = 5.6 × 10²³ kg.
1) 
When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude
where
q is the electron charge
E is the magnitude of the electric field
v is the electron speed
B is the magnitude of the magnetic field
Solving the formula for v, we find
2) 4.1 mm
When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:
where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find
3) 
The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, 
, and the period, T:
Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:
