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creativ13 [48]
3 years ago
11

An airplane flies 40km directly south in 10 minutes and 20 km directly east in five minutes, its average speed is:

Physics
2 answers:
blondinia [14]3 years ago
8 0
Average speed = total distance / total time
total distance = 40 + 20 = 60km
total time taken = 10 + 5 = 15 minutes
Average speed = 60/15 = 4km/min
Zina [86]3 years ago
3 0

Answer:

D. 4km/min

Explanation:

Since the question deals with "speed" and not "velocity", the direction does not matter.

the plane covers 40 + 20 km=60km in 10mins +5 mins=15mins

60 km - 15 mins

1min- 60÷15= 4 mins

therefore, your answer is D. 4km/min

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Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.
Alexus [3.1K]

Answer:

a) 9.00 m b) 6.00 m/s  c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = \frac{1}{2}*a*t^{2} = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

a = \frac{2*3.00m}{1s2} = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + \frac{1}{2}*a*t^{2} = 6.00m/s*1s +\frac{1}{2}*6.00 m/s2*1s^{2}  = 9.00 m

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s

3 0
3 years ago
a ball on a string is swung overhead in a horizontal circle with a radius of 0.66 m. what speed (in m/s) does the ball need to m
AnnZ [28]

Answer:

The ball is moving in a horizontal circle.

The force acting towards the center of the circle is

F = M a = M * V^2 / R    

a = V^2 / R      simplifying equation

V = (a / R)^1/2 = (23 / .66)^1/2 = 5.90 m/s

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2 years ago
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Zigmanuir [339]

Explanation:

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Refraction

Diffraction

4 0
3 years ago
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stiks02 [169]

Answer:

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Explanation:

8 0
3 years ago
A 1.0 kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s. After sliding a distance of 2.0 m, the bl
BARSIC [14]

Answer:

3.9 m

Explanation:

The principle of work and energy

ΔE = W  Formula (1)

where:

ΔE: mechanical energy change (J)

W : work of the non-conservative forces (J)

ΔE = Ef - E₀  

Ef : final mechanical energy

E₀   : initial mechanical energy

Ef = K f+ Uf

E₀ = K₀ + U₀

K =(1/2 )mv² :  Kinetic energy (J)

U = mgh  :Potential energy (J)

m: mass (kg)

v : speed (m/s)

h: hight (m)

Known data

m = 1 kg  : mass of the block

v₀ =  mg(h).0 m/s Initial speed of the block

vf = 0 = Final speed of the block

θ =40°  :angle θ of the ramp with respect to the horizontal direction

μk=0  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

 Problem development

W = 0 ,  Because the friction force (non-conservative force ) is zero

Principle of work and energy to the Block:

ΔE = W

Ef - E₀  = 0   Equation (1)

Ef = K f+ Uf =(1/2 )m(0)² + mg(h)=  mg(h)  (Joules)

E₀ = K₀ + U₀ = (1/2 )m(7)² + mg(0) = 24.5m (Joules)

In the equation (1) :

Ef =  E₀

mg(h) = 24.5m

We divided  by m both sides of the equation

g(h) = 24.5

h = 24.5 / g

h = 24.5 / 9.8

h = 2.5 m

We apply trigonometry at the ramp to calculate how far up the ramp (d) does the block slide before coming momentarily to rest :

sinθ = h/d

d = h / sinθ

d = 2.5 m / sin40°

d = 3.9 m

5 0
3 years ago
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