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sergij07 [2.7K]

3 years ago

7

A car is travelling at a speed of 90km/h. On seeing baby 20m ahead on the road, the driver jamed on the breakes and it came to r

edt at a distance of 15m. What is ots ratardation and how long does it take to come rest

1 answer:

Thepotemich [5.8K]3 years ago

6 0

Answer:

5m

Explanation:

because if the driver jamed the brakes at 20m and stop at 15m the it took the driver 5m to fully stop

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Properties of light that define light as a wave?

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The first choices are correct, because the second choices could happen by things other than light.

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A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

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Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

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(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

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x = 48 meters

Hence, this is the required solution.

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3 years ago

An airplane is flying in the direction 10° east of south at 701 km/hr. Find the component form ofthe velocity of the airplane, a

solniwko [45]

Answer:

The component form will be;

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Explanation:

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3 years ago

How to calculate the angle of a resultant<br>

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3 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

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Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

3 years ago