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4 years ago

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A graph of x vs. t2 is linear, and intercepts the vertical axis at 12 m and the horizontal axis at 4 s2. What is the function?

1 answer:

gogolik [260]4 years ago

4 0

Answer:

x = -3t² + 12

Explanation:

x vs t² is a line.

x = at² + b

The y intercept is 12.

x = at² + 12

At t² = 4s², x = 0.

0 = a(4) + 12

a = -3

Therefore, the function is:

x = -3t² + 12

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Suppose a star the size of our Sun, but of mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 17 days. If

Tanya [424]

Use the conservation of angular momentum; angular momentum at the beginning = angular momentum at the end
Conservation of angular momentum:
I1 w1 = I2 w2
Where I is the moment of inertia. For a sphere, I=2/5 m R^2. Substituting into the equation above we get
w2 = I1 w1 / I2 = w1 m1 R1^2 / (m2 R2^2)
w2 = w1 4 * (R1/R2)^2
= 4*(1)*(7E5/7.5)^2
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3 years ago

A car starting from rest moves with a constant acceleration of 10 mi/hr2 for 1 hour, then decelerates at a constant 5 mi/hr2 unt

Vinvika [58]

Answer:

The total distance traveled by the car (S) = 15 mi

Explanation:

Total acceleration ( $a_{1}$ ) = 10 mi / $h^{2}$

Time (t) = 1 hour

Deceleration ( $a_{2}$ ) = - 5 mi / $h^{2}$

From law of speed we know that V = u + $a_{1}$ t -----------(1)

Where V = Final speed

u = initial speed and t = time period

Car starts from rest so initial speed is zero so u = 0

Then equation 1 becomes V = $a_{1}$ t = 10 × 1 = 10 mi / h ---------(2)

The distance traveled by the car $S_{1}$ = u t + $\frac{1}{2}$ × $a_{1}$ × $t^{2}$ -------------(3)

Put all the values in equation we get $S_{1}$ = 0 + $\frac{1}{2}$ × 10 × $1^{2}$

$S_{1}$ = 5 mi ---------------(4)

This is the distance traveled by the car in 1 hour.

In second case when car starts decelerates the equation of motion becomes

⇒ V = u - a t -----------(5)

final speed in this case = 0 and initial speed (u) = 10 mi / h

⇒ 0 = 10 - 5 t

⇒ 5 t = 10

⇒ t = 2 hours

Distance traveled by the car in to hours $S_{2}$ = u t + $\frac{1}{2}$ × $a_{2}$ × $t^{2}$

⇒ $S_{2}$ = 10 × 2 - $\frac{1}{2}$ × 5 × $2^{2}$

⇒ $S_{2}$ = 20 - 10

⇒ $S_{2}$ = 10 mi

This is the distance traveled by the car in the next two hours.

Thus the total distance traveled by the car = $S_{1}$ + $S_{2}$

⇒ S = 5 + 10 = 15 mi

3 0

3 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

patriot [66]

If I have done my math correctly, your answer is 67.5

4 0

3 years ago

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A car travels at a constant speed of 40 miles per hour down the street for around 0.25 hours. What is its acceleration?

forsale [732]

It’s acceleration is 0 if it was traveling at a constant speed

5 0

3 years ago

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A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction

anyanavicka [17]

Answer:18.5 kg

Explanation:

Given

diameter $d=75 cm$

mass of stone $m=3 kg$

velocity of stone $v=3.4 m/s$

height fallen $h=2.40 m$

using

$v^2-u^2=2as$

$(3.4)^2=2\times a\times 2.4$

$a=2.4 m/s^2$

Tension of string will Provide Torque to cylinder

$T\times r=I\times \alpha$

where $I=moment\ of\ inertia$

$\alpha =angular\ acceleration$

$T=\frac{Mr^2}{2}\times \frac{a}{r}$

$T=\frac{Ma}{2}$

and $mg-T=ma$

$mg-ma=T$

Put value of T

$mg-ma=\frac{Ma}{2}$

$mg=a(m+\frac{M}{2})$

$3\times 9.8=2.4\cdot (m+\frac{M}{2})$

$12.25=3+\frac{M}{2}$

$9.25=\frac{M}{2}$

$M=18.5 kg$

4 0

4 years ago