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Yuri [45]
3 years ago
14

How long will it take for the car to accelerate

Physics
1 answer:
solniwko [45]3 years ago
8 0
You have to be 90 m away from the cat

Just divide 90 by 17.4 and you will get - 5.172 seconds

- It will take 5.17 seconds to accelerate uniformly to stop in exactly 2 m before the cat
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Pierre cycles at 4 miles per hour and travels a distance of 13 miles. How long
amm1812

Answer:

His journey took him 3 hours 15 minutes.

Explanation: 4 miles every hour. So 1 hr is equal to 4 miles, 2 hrs is equal to 8 miles, 3 hrs is equal to 12 miles. Now he just has 1 miles left, and since it takes him a hour to cycle 4 miles, 60 divided by 4 is 15. Therefore, 1 mile is equal to 15 minutes.

4 0
4 years ago
Read 2 more answers
Chich expression correctly describes force using Sl units
Zinaida [17]

Answer:1 N = 1 kg·m/s2

Explanation:

5 0
3 years ago
A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina
elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec        

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car a=1.6m/sec^2 in negative direction so acceleration will be a=-1.6m/sec^2

From first equation of motion we know that

v = u+at

So v=0+(-1.6)\times 5.8=-9.28m/sec

So final velocity will be -9.28 m/sec

8 0
4 years ago
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas
timurjin [86]

Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

          = 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

          = 0.135 J + 0.11220 J

          = 0.2472 J

          ≅ 0.247 J = 247 mJ

5 0
3 years ago
At temperatures near absolute zero, Bc approaches 0.142 T for vanadium, a type-I superconductor. The normal phase of vanadium ha
ioda

Answer:

b) field is zero,  c) the magnetic field does not change in intensity or direction

e) M = -H = Bo /μ₀ ,  g)  M = 0

Explanation:

Part b

superconductors are formed by so-called Coper pairs that are electrons linked through a distortion in the network, this creates that they must be treated as an entity so we have an even number of charge carriers and the material must behave with diamagnetic , Meissner effect, consequently the magnetic field inside its superconductor is zero

the correct answer is Zero

Part c

 outside the superconducting cylinder the magnetic field does not change in intensity or direction

Part E

Magnetization is defined by the equation

       B = μ₀ (H + M)

with field B it is zero inside the superconductors

        M = -H = Bo /μ₀

         

where Bo is the magnetic induction in the normal state

Part g

 As outside the cylinder there is no material zero magnetization

        M = 0

6 0
3 years ago
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