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Setler [38]
3 years ago
9

Cross country skier moves 32 m west word than 54 m east word and finally 68 m westward what is the distance moved?

Physics
1 answer:
Juliette [100K]3 years ago
8 0

Answer:

154m

Explanation:

Distance moved

= 32m + 54m + 68m

= 154m

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Can we use a clinical thermometer to measure the temperature of a candle flame​
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Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a
seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

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s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

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s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

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3 years ago
A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
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3 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

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From the law of conservation of linear momentum

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Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

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V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
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