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Marianna [84]
3 years ago
9

What was the average speed in km/h of a car that travels 400.0 km in 4.5 h?

Physics
1 answer:
Yanka [14]3 years ago
6 0
V= s/t = 400/4.5=800/9 (km/h)
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A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.
Lilit [14]

Answer:

Explanation:

a. The source of centripetal force on the car is  (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c.  Fc = m(ac) = 1500(8) = 12 kN

d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

6 0
3 years ago
What is a group of individuals of the same species living in the same geographic area
balu736 [363]
A group of individuals  living in a particular geographic area is termed population.
8 0
3 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
Is it possible to apply the same amount of force and do different amounts of work?
Tamiku [17]
Yes, with simple machines
5 0
3 years ago
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