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klio [65]
2 years ago
15

A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car

slams on her brakes and skids on dry pavement for 50 m , decelerating at 7.5 m/s2 . Then she hits the icy patch and skids another 70 m before coming to rest.
A) If her initial speed was 78 mi/h , what was the magnitude of the deceleration on the ice?
Physics
1 answer:
dangina [55]2 years ago
5 0
Her magnitude of deceleration on the ice would be 15.126m/s
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What do measurements of the number of neutrinos emitted by the Sun tell us about conditions deep in the solar interior?
OverLord2011 [107]

Answer:

The neutrinos are produced in the core of the sun by nuclear fusion and measuring their number helps us confirm that there are enough proton-proton chain reactions of each which produce a neutrino and going on in the Sun's core to explain the energy output of the Sun.

4 0
2 years ago
A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up
Andrew [12]

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

6 0
3 years ago
Question 10(Multiple Choice Worth 7 points)
nataly862011 [7]

Organisms that existed over a limited period

Explanation:

Index fossils are organisms that existed within a limited period in the geologic record.

These fossils are indicator fossils that serves as guide to define the relative ages of rocks in a sequence.

Here are some characteristics of index fossils:

  • they must have existed over a limited period in their history.
  • they must be very distinctive with short evolutionary tend.
  • they must be abundant and highly widespread in the stratigraphic column.

Learn more:

Fossils brainly.com/question/7382392

#learnwithBrainly

5 0
3 years ago
Which type of ooze dominates in warm shallow ocean floors?<br>A. Calcareous<br>B. Silieous
gogolik [260]
Actually it's sediment
6 0
3 years ago
It's a snowy day and you're pulling a friend along a
NeX [460]

The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

<h3>How to find the coefficient of friction between the sled and the snow?</h3>
  • Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.
  • To solve the problem, we have to draw the free body diagram of the given system.
  • We have given with the following values,

                                     a=0\\\alpha =35^0\\T=75N\\m=57kg

Here, acceleration will be equal to zero, because the velocity is given as constant.

  • Thus, from the diagram, we can write the balancing equations as follows,

                                      ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha

  • Substituting N in f and f in the equation of ma, then we get,

                   ma= Tcos\alpha -k(mg-Tsin\alpha )

  • Substituting values, we get the coefficient of friction as,

                    0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119

Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.

Learn more about the friction here:

brainly.com/question/28107059

#SPJ1

6 0
1 year ago
Read 2 more answers
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