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Rama09 [41]
2 years ago
10

scientific __________ involves the knowledge and understanding necessary for evaluating information, making personal decisions,

and taking part in public affairs
Physics
2 answers:
mash [69]2 years ago
4 0

Answer;

-Scientific literacy

Explanation;

-Scientific Literacy is the knowledge of understanding of scientific terms and principles required for evaluating information, making personal decisions, and taking part in public affairs. By having this, one can be able to identify good sources of scientific information, evaluate them for accuracy, and apply them for knowledge to questions or problems in their life.

-It is the knowledge and understanding of scientific concepts and processes required for personal decision making, participation in civic and cultural affairs, and economic productivity. It also includes specific types of abilities. A society with a higher science literacy would be able to make better judgments and decisions.

Alenkinab [10]2 years ago
4 0

Answer:

litearcy

Explanation:

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. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?
balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

U = K

\frac{kQe}{R} = \frac{1}{2}mv^2

\frac{2kQe}{mR} = v^2

now take square root of both sides

v =\sqrt{\frac{2kQe}{mR}}

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

k = 9 * 10^9


5 0
2 years ago
Read 2 more answers
Pls help it's due today​
mario62 [17]

Answer:

Breh seriously. Ugh fine.

1.B

2.D

3.C

4.C

5.D,A and B

6.A,C and D

3 0
2 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
2 years ago
Newton's _____ law explains why my hands hurt when I clap loudly
Basile [38]
Newtons third law (inertia) is to blame
3 0
3 years ago
Read 2 more answers
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