In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
Answer:
B = 62.9 N
Explanation:
This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid
B = ρ g V
write the equilibrium equation
T + B -W = 0
B = W- T (1)
use the density to write the weight
ρ = m / V
m = ρ V
W = ρ g V
substitute in 1
B = m g -T
B = 
g V - T
To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³
calculate
B = 7850 9.8 1.20 10⁻³ - 29.4
B = 92.3 - 29.4
B = 62.9 N
The distance of the canoeist from the dock is equal to length of the canoe, L.
<h3>
Conservation of linear momentum</h3>
The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.
v(m₁ + m₂) = m₁v₁ + m₂v₂
where;
v is the velocity of the canoeist and the canoe when they are together
- u₁ is the velocity of the canoe
- u₂ velocity of the canoeist
- m₁ mass of the canoe
- m₂ mass of the canoeist
<h3>Distance traveled by the canoeist</h3>
The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.
Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.
Learn more about conservation of linear momentum here: brainly.com/question/7538238
Answer:
(a) 1.85 m/s
(b) 4.1 m/s
Explanation:
Data
- initial bullet velocity, Vbi = 837 m/s
- wooden block mass, Mw = 820 g
- initial wooden block velocity, Vwi = 0 m/s
- final bullet velocity, Vbf = 467 m/s
(a) From the conservation of momentum:
Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf
Mb*(Vbi - Vbf)/Mw = Vwf
4.1*(837 - 467)/820 = Vwf
Vwf = 1.85 m/s
(b) The speed of the center of mass speed is calculated as follows:
V = Mb/(Mb + Mw) * Vbi
V = 4.1/(4.1 + 820) * 837
V = 4.1 m/s
To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m
