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3 years ago

13

A car has a uniform velocity of 108km/h. How far does it travel in 30 seconds?

1 answer:

natita [175]3 years ago

6 0

Velocity=108km/h

Convert to m/s

$\\ \tt\longmapsto 108\times \dfrac{5}{18}$

$\\ \tt\longmapsto 30m/s$

Time=30s.

$\\ \tt\longmapsto Distance=Velocity\times Time$

$\\ \tt\longmapsto Distance=30(30)$

$\\ \tt\longmapsto Distance=900m$

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50 mL of soda in a soda can exerts ____ 50 mL of soda in a 1L bottle.

lutik1710 [3]

More pressure than..

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3 years ago

An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m

solong [7]

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

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8 0

3 years ago

You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba

Cloud [144]

Answer:

The ball reaches Barney head in $t = 8 \ s$

Explanation:

From the question we are told that

The rise velocity is $v = 14.70 \ m/s$

The height considered is $h = 196 \ m$

The horizontal velocity of the large object is $v_h = 8.50 \ m/s$

Generally from kinematic equation

$s = ut + \frac{1}{2} gt^2$

Here s is the distance of the object from Barney head ,

u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So

$u = -14.7 m/s$

So

$196 = -14.7 t + \frac{1}{2} * 9.8 * t^2$

= $4.9 t^2 - 14.7t - 196 = 0$

Solving the above equation using quadratic formula

The value of t obtained is $t = 8 \ s$

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3 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

So

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

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3 years ago

Complete the equation???

soldi70 [24.7K]

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3 years ago