Answer:B
Explanation:
Given
Distance of astronaut From asteroid x is 
Distance of astronaut From asteroid Y is 
Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y
If the astronaut is in equilibrium then net gravitational force on it is zero


cancel out the common terms we get




Answer:
a = 4.72 m/s²
Explanation:
given,
mass of the box (m)= 6 Kg
angle of inclination (θ) = 39°
coefficient of kinetic friction (μ) = 0.19
magnitude of acceleration = ?
box is sliding downward so,
F - f = m a
f is the friction force
m g sinθ - μ N = ma
m g sinθ - μ m g cos θ = ma
a = g sinθ - μ g cos θ
a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°
a = 4.72 m/s²
the magnitude of acceleration of the box down the slope is a = 4.72 m/s²
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
Answer:
Q=1005 J
t= 0.67 sec
Explanation:
Lets take condition of room is 1 atm and 25°C.
Heat capacity ,c = 21 J /K.mol
If we assume that air is ideal gas that
P V = n R T



V= 107250 L
At STP number of moles given as

V=22.4 L at S.T.P.

n=4787.94 moles
n= 4.784 Kmoles
So heat required to raise 10°C temperature
Q = n x c x ΔT
Q = 4.78794 x 21 x 10
Q=1004.64 J
Time t
t= Q/P
P= 1.5 KW
t = 1.004.64 /1.5
t= 0.66 sec
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²