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Alika [10]
3 years ago
11

Need physics help! (there’s a picture)

Physics
1 answer:
Nezavi [6.7K]3 years ago
5 0

Answer:

i. Subtract v₀² from both of the equation.

ii. Divide both side of the equation by 2

iii. Divide both side of the equation by (x – x₀)

a = (v² – v₀²) / 2(x – x₀)

Explanation:

v² = v₀² + 2a(x – x₀)

To solve for a in the expression above, do the following:

i. Subtract v₀² from both of the equation.

v² – v₀² = v₀² – v₀² + 2a(x – x₀)

v² – v₀² = 2a(x – x₀)

ii. Divide both side of the equation by 2

(v² – v₀²) /2= 2a(x – x₀)/2

(v² – v₀²) /2 = a(x – x₀)

iii. Divide both side of the equation by (x – x₀)

(v² – v₀²) /2 ÷ (x – x₀) = a(x – x₀)/(x – x₀)

(v² – v₀²) /2 × 1/ (x – x₀) = a

a = (v² – v₀²) / 2(x – x₀)

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Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

V=\frac {1}{3}\times \pi\times r^2\times h

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

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A student measures the speed of a rolling ball three times. She adds the measurements and divides by 3. What quantity did the st
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The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
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Please help!! The graph in the figure shows the position of a particle as it travels along the x-axis. What is the magnitude of
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Answer:

1.3 m/s

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6 0
3 years ago
A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.
dalvyx [7]

Answer:

(a) W_c=127.008 J

(b) W_g=148.176 J

(c) K.E. = 21.168 J

(d) v=3.4293m.s^{-1}

Explanation:

Given:

  • mass of a block, M = 3.6 kg
  • initial velocity of the block, u=0 m.s^{-1}
  • constant downward acceleration, a_d= \frac{g}{7}

\Rightarrow That a constant upward acceleration of \frac{6g}{7} is applied in the presence of gravity.

∴a=- \frac{6g}{7}

  • height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

F_c= M\times a

F_c=3.6\times (-6)\times\frac{9.8}{7}

F_c=-30.24 N

∴Work by the cord on the block,

W_c= F_c\times d

W_c= -30.24\times 4.2

We take -ve sign because the direction of force and the displacement are opposite to each other.

W_c=-127.008 J

(b)

Force on the block due to gravity:

F_g= M.g

∵the gravity is naturally a constant and we cannot change it

F_g=3.6\times 9.8

F_g=35.28 N

∴Work by the gravity on the block,

W_g=F_g\times d

W_g=35.28\times 4.2

W_g=148.176 J

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

K.E.= W_g+W_c

K.E.=148.176-127.008

K.E. = 21.168 J

(d)

From the equation of motion:

v^2=u^2+2a_d\times d

putting the respective values:

v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }

v=3.4293m.s^{-1} is the speed when the block has fallen 4.2 meters.

6 0
3 years ago
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