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galben [10]
2 years ago
6

An airplane which intends to fly due south at 250 km/hr experiences a wind blowing westward at 40 km/hr. What is the actual spee

d of the airplane relative to the ground? ​
Physics
1 answer:
sleet_krkn [62]2 years ago
7 0

Answer:

simple is rumple a daily ok I'll be

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In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

4 0
3 years ago
which fire extinguisher is most appropriate to put out a fire that involves a stack of burning newspapers
evablogger [386]

Monoammonium phosphate effectively smothers the fire, while sodium bicarbonate induces a chemical reaction which extinguishes the fire. Fire extinguishers with a Class C rating are suitable for fires in “live” electrical equipment.

6 0
2 years ago
Read 2 more answers
Usain Bolt accelerates at a rate of 3.7
MissTica

Answer:

Explanation:

I think that you to run more than 12 miles

4 0
3 years ago
A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of t
emmasim [6.3K]

Answer:

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

(3 × g × cos 30°) ÷ 2× L

Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²

5 0
3 years ago
PLS ANSWER FAST WILL GIVE BRAINLEST!!!
Burka [1]

Answer:

F = 6,000 N

Explanation:

F = m × a

F = 2,000 × 3

F = 6,000 kg m/s²

F = 6,000 N

7 0
3 years ago
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