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djverab [1.8K]
2 years ago
7

Please help

Physics
2 answers:
vladimir2022 [97]2 years ago
5 0

Answer:

true true false true false

Greeley [361]2 years ago
4 0

Answer:

1)true

2)true

3)false

4)false

5)true

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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
The three osicles known as the hammer, anvil and stirrup __________.
34kurt
Malleus, incus, and stapes, respectively, and collectively, as "middle ear ossicles<span>".</span>
3 0
3 years ago
A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.
Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

a=\frac{v^{2} }{r}, where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

a=\frac{3.5^{2} }{8.5} \\a=1.4

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

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d1i1m1o1n [39]
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Dennis_Churaev [7]
I THINK C BECAUSE IF IT IS A GLASS BOX HOW DID A CACTUS GET IN AND NOTHING CAN GET IN OR OUT OF THE BOX SO THERE IS NO CACTUS IN THE BOX
6 0
3 years ago
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