Answer:
Explanation:I don't say you must have to mark my ans as brainliest but if you think it has really helped u plz don't forget to thank me....
The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:
2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>
Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) =
=
= 392 J
(C) average frictional force = 
- change in KE (ΔKE) = initial KE - final KE
- ΔKE =
-
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE =
- 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] =
=
= 212.33 N
Well first of all, the Space Shuttle program ended a few years ago, and none have been launched since then.
The Shuttle never went to places that were properly referred to as "outer space". When they flew, the Space Shuttles went to low Earth orbit, where the acceleration of gravity is roughly 85% of its value on the Earth's surface.
So a Shuttle that weighed 20 million Newtons on the launch pad weighed roughly 17 million Newtons while in orbit.
Answer:
The impuise is 7.9905 kg*m/s
Explanation:
Step one:
given data
v1= +2.63m/s
v2=-20.2m/s
mass m= 0.350kg
Step two:
From the expression for impulse
Ft= mΔv
substituting our data into the expression we have
Ft= 0.35*(-20.2-2.63)
Ft= 0.35*22.83
Ft=7.9905 kg*m/s