1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
FromTheMoon [43]
3 years ago
6

Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s

top in 70 feet. If the van is traveling 48 miles per​ hour, what is its stopping​ distance?
Physics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

v_f^2 - v_i^2 = 2 ad

so here we have

0^2 - 40^2 = 2 a (70 feet)

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

0^2 - 48^2 = 2 a (d)

now divide the above two equations

\frac{40^2}{48^2} = \frac{70 feet}{d}

d = 100.8 ft

You might be interested in
four forces act on an object of mass 100kg, they are 12N(N), 70N(S), 33N(E), 10N(w). calculate the reseulting acceleration of th
trapecia [35]

Answer:

0.62 m/s² at 68° S of E

Explanation:

Net force north = 12 - 70 = -58 N

Net force east = 33 - 10 = 23 N

Net force = √(-58² + 23²) = 62.3939... N

acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²

θ = arctan(-58/23) = -68.3691... ≈ 68° S of E

6 0
3 years ago
Wich of the following is NOT an example of accelerated motion?
Allushta [10]
B is the answer for your question the answer is b an elevator slowing down  
5 0
3 years ago
Read 2 more answers
On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s
Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0
3 years ago
it took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 miles per hour. How many miles lie
Art [367]

Given:

Time: 3.5 hrs

Velocity: 120 miles/hr

Now Distance=  Speed × Time

Now Velocity and speed have the same magnitude. Velocity being a vector quantity has a definite direction. Whereas speed is a scalar quantity,it indicates only the magnitude an doesn't define any direction.

Hence Distance = Velocity x time

Distance = 3.5 × 120 = 420 miles

7 0
3 years ago
How many earths could fit inside jupiter
Snezhnost [94]

1,300 or more.

hope this helped :)

8 0
3 years ago
Other questions:
  • Calculate the density of a sample of gas with a mass of 30g and volume of 7500 cm3
    11·1 answer
  • a driver brings a car to a full stop in 2.0 s. if the car was initially traveling at 22 m/s, what is the acceleration?
    9·1 answer
  • Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis t
    12·1 answer
  • Dormancy helps plants survive freezing temperatures and lack of..
    8·1 answer
  • A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil
    14·1 answer
  • The earth's orbital is oval in shape. Explain how the magnitude of the gravitational force between the earth and the sun changes
    14·1 answer
  • FLAMING OR ALBERT .........
    14·2 answers
  • PLEASE SEE MY QUESTIONS UNDER MY PROFILE AND ANSWER THEM, I REALLY NEED HELP THANK YOU!
    11·2 answers
  • You see the moon rising, just as the sun is setting. What phase is the moon in?.
    10·1 answer
  • A thermal reservoir can be characterized as a thermal body that is large enough that when energy is dumped into it or taken out
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!