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ANEK [815]
3 years ago
10

What is the maximum density of water​

Physics
1 answer:
Reika [66]3 years ago
5 0

Answer:

4ºC or 39.2ºF

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4) A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it’s moving at 53 m/s at the end of this interval
Scorpion4ik [409]

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

    D = \dfrac{v_0 + v_1}{2}\times t

    140 = \dfrac{v_0 + 53}{2}\times 3.6

       v₀ + 53 = 77.78

        v₀ = 24.78 m/s or 25 m/s

b) a = \dfrac{v-u}{t}

   a = \dfrac{53-25}{3.6}

         a = 7.8 m/s²

using equation of motion

  v₀² = v₁² + 2 a s

  53² = 0²+ 2 x 7.8 x s

  s = 180 m

3 0
3 years ago
Are amplitude and wavelength related?<br> A: Yes<br> B: No
Trava [24]

The answer is A.Yes

Explanation:

The amplitude of a wave is the height of a wave as measured from the highest point of the wave to the lowest on the wave.

4 0
2 years ago
One of the light bulbs in this series circuit burns out, causing a break in the circuit.
ivanzaharov [21]

Answer:

It doesn't give light

Explanation:

No Flowing of electricity

3 0
2 years ago
How does Piggy feel about being called "Piggy"?
denpristay [2]

Answer:

good because piggy loves peppa pig :3

Explanation:

4 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
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