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Greeley [361]
3 years ago
13

A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an

d the mass of the sled is 10.3 kg.
The force of friction acting on the sled as it slides is -123 N.

If the sled was initially moving at 13.3 m/s at t=0, how far will it slide before the sled stops?
Physics
1 answer:
jonny [76]3 years ago
6 0

Answer:

28.81 m

Explanation:

Ff = -123

m * a  = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

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Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

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make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

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v = √(1742)

v = 41.74 m/s.

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Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res
MakcuM [25]

Answer:

The mass of the beam is 0.074 kg

Explanation:

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Set up this system with the given mass and support;

  0-----------------33cm-----------------------------------100cm

  ↓                     Δ                                             ↓      

0.15kg                                                              m

Where;

m is mass of the uniform bar

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