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3 years ago

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A rocket powered sled is accelerating along a straight, level track with a constant acceleration of magnitude 20 m/s2. By how mu

ch will the speed of the sled change in 3.00 seconds

1 answer:

Dafna1 [17]3 years ago

7 0

Answer:

the change in speed of the sled after 3.0 s is 60 m/s.

Explanation:

Given;

constant acceleration of the sled, a = 20 m/s²

change in time of motion, dt = 3.0 s

The change in speed of the sled after 3.0 s is calculated as;

$a = \frac{dv}{dt} \\\\dv = a \times dt\\\\dv = 20 \ \times \ 3\\\\dv = 60 \ m/s$

Therefore, the change in speed of the sled after 3.0 s is 60 m/s.

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Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope

VladimirAG [237]

Car is moving on the glassy slope with constant speed

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$a = \frac{dv}{dt}$

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

$a = 0$

now as we know by Newton's II law

$F = ma$

since a = 0

$F = 0$

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3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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3 years ago