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Vlad [161]
3 years ago
7

Dos cargas muy pequeñas de masas iguales m

Physics
1 answer:
NemiM [27]3 years ago
3 0

Answer:

b)A partir de dicha expresión encontrar el ángulo a en

función de m, L y q cuando a < 5º.

Explanation:

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be 8.21\times 10^3~N

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0
3 years ago
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
White raven [17]

Answer:

400ft.    32ft/s       -32ft/s

Explanation:

In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2

Anyway for the sake of assumtion let us takes=160t-16t^2

 

ds/dt=160-32t=0

 

t=160/32= 5 seconds.

s=160*160/32-16*(160/32)^2= 400 mts

 

 

s=384 mts

160t-16t^2=384

i.e

16t^2-160t+384=0

 

t^2-10t+24=0

(t-6)(t-4)=0

t=[4,6]

we have to take t=4 because it is all the up i.e <5

 

velocity =v=ds/dt=160-32t

 

v=160-32*4=32 ft/sec still going up

 

for all the way down take t=6 whuch is >5

 

v=160-6*32=-32 ft/sec (falling down!!!)

6 0
3 years ago
Name 5 ways you can deal with a bully without engaging in a physical altercation with him/her.
Delicious77 [7]

Answer:

ignore the bully ,tell the bully to stop,make a joke or laugh with the bully,stick with friend,know how to get out of the bullying situation

Explanation:

6 0
3 years ago
If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

3 0
3 years ago
Which test was conducted on the skin found under the victim’s nails?
BARSIC [14]
Y-STR analysis. It is a DNA test used to compute and match with the possibility via lap.
5 0
3 years ago
Read 2 more answers
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