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zavuch27 [327]
3 years ago
10

The stoplight had just changed and a 2200 kg cadillac had entered the intersection, heading north at 2.8 m/s , when it was struc

k by a 1200 kg eastbound volkswagen. the cars stuck together and slid to a halt, leaving skid marks angled 35∘ north of east.
Physics
1 answer:
tekilochka [14]3 years ago
7 0
<span>3. The attempt at a solution So basically what I did was divided into components. x: (3)(2000) = (3000)*v_x y: (v_vw)*(10000) = (3000)*v_y v_x, v_y is the velocity (after collision) in the x and y direction, respectively, of both cars stuck together (since it is an inelastic collision). v_vw is the initial velocity of the Volkswagen. Now what I did was that the angle is 35 degrees north of east. So basically made a triangle and figured that tan(35) = (v_y)/(v_x). This means (v_x)*(tan35) = v_y. Then, I simplified the component equations to get: x: 2 = v_x y: v_vw = 3*v_y Then plugging in for v_y, I got: v_vw = 3(2)(tan35) = 4.2 m/s as the velocity of the volkswagen. However, the answer key says 8.6 m/s. Could someone please help me out? Thanks Phys.org - latest science and technology news stories on Phys.org • Game over? Computer beats human champ in ancient Chinese game • Simplifying solar cells with a new mix of materials • Imaged 'jets' reveal cerium's post-shock inner strength Oct 24, 2012 #2 ehild Homework Helper Gold Member What directions you call x and y? Reference https://www.physicsforums.com/threads/2d-momentum-problem.646613/</span>
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How much salt (nacl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/l of salt? give your answer in kg/day?
slamgirl [31]
Answer: 129,600kg/day
In this question, you are given the river flow (30m3/s) and the salt concentration. Then to determine the amount of salt carried, you simply need to multiply the flow with concentration. Be careful because the unit in this question is different and need to be converted. The equation would be:

Salt = river flow x salt concentration
Salt = 30m3/s x 1000L/m3 x 50mg/L = 1,500,000mg/second

Then convert it into kg/day
Salt= 1,500,000mg/second x 10^-6 kg/mg x (3600 second/hour) x (24 hour/day)= 129,600kg/day
7 0
3 years ago
Which learning style describes children who use the body effectively like a surgeon or a dancer?
creativ13 [48]

Answer:

A. Bodily Kinesthetic

Explanation:

3 0
3 years ago
The battery charger for an mp3 player contains a step-down transformer with a turns ratio of 1:38, so that the voltage of 120 v
matrenka [14]
Transformer contains two coils: primary and secondary. They allow change of voltage to lower or higher value. In first case we have step-down and in second case we have step-up transformer.
Formula used for transformer is:\frac{N_{1} }{N_{2}} = \frac{V_{1}}{V_{2}}

Where:N1 = number of turns on primary coilN2 = number of turns on secondary coilV1 = voltage on primary coilV2 = voltage on secondary coil
In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:38 means that for each turn on secondary coil we have 38 turns on primary coil.
We can solve the equation for V2:V_{2} =  \frac{ V_{1}* N_{2}  }{ N_{1} }  \\  V_{2} =  \frac{ 120* x  }{ 38x} } \\ V_{2} = 3.16V

Secondary coil provides voltage of 3.16V.
7 0
4 years ago
How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl
Vinvika [58]

Answer:

B) \frac{2}{3}

Explanation:

The electric force between charges can be determined by;

F = \frac{kq_{1} q_{2} }{r^{2} }

Where: F is the force, k is the Coulomb's constant, q_{1} is the value of the first charge, q_{2} is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

q_{1} = 2q

q_{2} = \frac{q}{3}

So that,

F = q_{1}q_{2} x \frac{k}{r^{2} }

  = 2q x \frac{q}{3} x \frac{k}{r^{2} }

  = \frac{2q^{2} }{3} x \frac{k}{r^{2} }

  = \frac{2}{3} x \frac{kq}{r^{2} }

F = \frac{2}{3} x \frac{kq}{r^{2} }

The electric force between the given charges would change by \frac{2}{3}.

4 0
3 years ago
A 2130 kg car is parked on a hill that makes a 15º angle with the horizontal. What is the normal force on the car?
valentinak56 [21]

Answer:

20573.67N

Explanation:

Given;

mass (m) of the car = 2130kg

angle of inclination Θ = 15⁰

The normal force (F) on the car is given by

F = mgcosΘ

where g is the acceleration due to gravity.

Taking g as 10m/s^{2} and substituting the values of m and Θ into the equation. We have;

F = 2130 x 10 x cos 15⁰

F = 2130 x 10 x 0.9659

F = 20573.67N

Therefore the normal force on the car is 20573.67N

7 0
4 years ago
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