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lakkis [162]
2 years ago
10

I WILL MARK YOU IF YOU HELP

Physics
2 answers:
aliya0001 [1]2 years ago
8 0
Try to get the best out of the way and the second is a good
sertanlavr [38]2 years ago
6 0
F=ma
11.6=3.8*a
a=11.6/3.8
a=3.05m/s
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Draw an energy chain diagram to show energy transformations for this event:
larisa86 [58]

The energy chain diagram in this case is chemical energy >> energy conversion >> motion (kinetic) energy + thermal energy.

<h3>What is an energy chain diagram?</h3>

An energy chain diagram is a graphic representation indicating the conversion between different types of energies.

Kinetic energy is a type of motion (movement) energy generated by using potential (stored) energy.

Chemical energy (in this case, the fuel of the car) is a type of energy that is stored to perform work.

Learn more about kinetic energy here:

brainly.com/question/25959744

8 0
2 years ago
Please help fast!!
daser333 [38]

The answer is C..........      

4 0
3 years ago
The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

5 0
3 years ago
A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 19
Hunter-Best [27]

The car will take 300 m before it stops due to applying break.

<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
  • As per Newton's equation of motion, V² - U² = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
  • So, 0² - 60² = 2×6× S

=> -3600 = -12S

=> S = 3600/12 = 300 m

Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

#SPJ1

7 0
1 year ago
g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron
Butoxors [25]

Answer:

The value is v  =  2.999 *10^{8} \  m/s

Explanation:

From the question we are told that

   The time taken to travel to the planet from earth is t = 189 \ light-years

    The  time to be spent on the ship is  t_{s} =  12 \  years

Generally speed can be obtained using the mathematical relation represented below

       t_s  =  2 * t *  \sqrt{1 -  \frac{v^2}{c^2 } }

The 2 in the equation show that the trip is a round trip i.e going and coming back

=>    12 =  2 * 189 *  \sqrt{1 -  \frac{v^2}{(3.0*10^{8})^2 } }

=>     v  =  2.999 *10^{8} \  m/s

5 0
3 years ago
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