Answer:

Explanation:
<u>Given Data:</u>
Momentum = P = 700 kg m/s
Velocity = v = 10 m/s
<u>Required:</u>
Mass = m = ?
<u>Formula:</u>
P = mv
<u>Solution:</u>
m = P / v
m = 700 / 10
m = 70 kg
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AnonymousHelper1807</h3>
Answer:
5.5 × 10^14 Hz or s^-1
no orange light has less frequency so no photoelectric effect
Explanation:
hf = hf0 + K.E
HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s
f is frequency of incident photon and f0 is threshold frequency
hf0 = hf- k.E
6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20
6.63 × 10 ^-34 × f0 = 3.64158×10^-19
f0 = 3.64158×10^-19/ 6.63 × 10 ^-34
f0 = 5.4925 × 10^14
f0 =5.5 × 10^14 Hz or s^-1
frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light
r = radius of the circle of the ride = 3.00 meters
v = linear speed of the person during the ride = 17.0 m/s
m = mass of the person in angular motion in the ride
L = angular momentum of the person in the ride = 3570 kg m²/s
Angular momentum is given as
L = m v r
inserting the values
3570 kg m²/s = m (17 m/s) (3.00 m)
m = 3570 kg m²/s/(51 m²/s)
m = 7 kg
hence the mass comes out to be 7 kg
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
Answer:
P_(pump) = 98,000 Pa
Explanation:
We are given;
h2 = 30m
h1 = 20m
Density; ρ = 1000 kg/m³
First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,
Thus, it can be expressed as;
P_(tank)+ P_(pump) = P_(nozzle)
Now, the pressure would be given by;
P = ρgh
So,
ρgh_1 + P_(pump) = ρgh_2
Thus,
P_(pump) = ρg(h_2 - h_1)
Plugging in the relevant values to obtain;
P_(pump) = 1000•9.8(30 - 20)
P_(pump) = 98,000 Pa