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masya89 [10]
3 years ago
13

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.7

5 m/s^2 for 4.40 s, making straight skid marks 60.0 m long, all the way to the tree. With what speed does the car then strike the tree? m/s
Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

0.99 m/s

Explanation:

t = Time taken = 4.4 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 60 m

a = Acceleration = -5.75 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{60-\frac{1}{2}\times -5.75\times 4.4^2}{4.4}\\\Rightarrow u=26.29\ m/s

v=u+at\\\Rightarrow v=26.29+(-5.75)\times 4.4\\\Rightarrow v=0.99\ m/s

The car hits the tree at 0.99 m/s

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Answer:

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X-direction                                     | Y-direction

x=x_o+v_{xo}t ⇒ x=v_{xo}cos(delta)t |

5 0
3 years ago
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 4
alexgriva [62]

Answer:

Number of turns on the secondary coil of the adapter transformer is 10.

Explanation:

For a transformer,

    \frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }

where V_{s} is the voltage induced in the secondary coil

           V_{p} is the voltage in the primary coil

          N_{s} is the number of turns of secondary coil

         N_{p} is the number of turns of primary coil

From the given question,

    \frac{487*10^{-3} }{120} = \frac{N_{s} }{2464}

⇒    N_{s} = \frac{2462*487*10^{-3} }{120}

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5 0
3 years ago
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bearhunter [10]
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4 0
3 years ago
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he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1
meriva

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

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r⊥ is the distance between the applied force and axis

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r = 56.0N-m

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56.0N-m² = (145rad/s²) x I

The I would be:

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I = 56/145

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This is the moment of inertia.

Thank you!

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