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3 years ago

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What graph shape is this what does the snap tell you

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vekshin13 years ago

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The picture is hard to see but if you still need help message me

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Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) resp

Sophie [7]

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: $x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ , where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

<u>x-values:</u>

<u /> $x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}$ = $\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}$ = $\frac{81}{41}$ = 1.9756

<u>y-values:</u>

<u /> $y_{cm}$ = $\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}$ = $\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}$ = $\frac{-90}{41}$ = -2.1951

<u>center of mass:</u>

(1.9756, -2.1951)

6 0

3 years ago

Calculate accelerationof the following. calcutale the acceleration of josh rding his biycle in a straight line that speeds up fr

BartSMP [9]

Explanation:

Wooho Physics my favorite subject! And my 100th answer :)

We will use the formula

$\boxed{\tt{ Acceleration = \dfrac{v - u}{t}}}$

Where v is Final velocity
And u is Initial velocity.
t is time taken.

In this question :-

u is 4 m/s
v is 6 m/s
t is 5 seconds

Applying the formula:-

$\sf{ Acceleration = \dfrac{6 - 4}{5}}$

$\sf{ Acceleration = \dfrac{2}{5}}$

So, Acceleration is 0.4 m/s² is the answer.

Hope it helps :)

8 0

3 years ago

2. A car accelerates down the road. What is the "reaction" to the tires pushing on the road?

Artist 52 [7]

The answer would be friction

7 0

3 years ago

A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis

olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = $\dfrac{1200}{60}$ rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

$T = \dfrac{1}{f}$

T = 1/20 = 0.05 s

6 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

3 years ago