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1 year ago

I need help with problem C, finding the area of the accel. v time graph

Physics

1 answer:

erastova [34]1 year ago

7 0

c) To determine the velocity of the motion at the given time, consider that the velocity is also the area under the line of the graph a vs t.

Between t=0s and t = 2 s, the area under the line is:

A1 = 2 X 2 = 4

Then, until t = 2s the velocity is 4 m/s

The area between t = 2s and t = 4 s is:

A2 = 2 x 0 = 0

The area between t = 4s and t = 5s, the area is:

A3 = 2 x 1 = 2

Now, consider that the velocity is the sum of the previous areas, but taking areas on the negative y-axis as negative ones. Then, you have:

A = A1 + A2 - A3 = 4 + 0 - 2 = 2

Hence, the velocity is 2 m/s

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A straw placed in a clear glass of water appears to break at the water’s surface. Which explains this effect? A). The water is t

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It is C. refraction causes this to happen.

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4 years ago

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1. A 17.45-N force is applied to a 3.10-kg object to accelerate it rightwards. The object encounters 15.25 N of

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3 years ago

A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear

vichka [17]

Answer:

a) λ = 5.19 10⁻⁴ C/m , b) E = 1,573 10⁻³ N/C , c) the direction of the field is directed to the bar

Explanation:

a) the linear density defined as the ratio between the charge per unit length

λ = q / l

Let's start by reducing the units to the SI system

L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

a = 12 cm (1m / 100cm) = 12 10⁻² m

q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

λ = 5.19 10⁻⁴ C/m

b) Let's look for the electric field for a point at a distance a from the end of the bar

E = k dq / r²

To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

λ = dq/dx

dq = λ dx

E = k ∫ λ dx / x²

We integrate and evaluate between the lower limits x = a and higher x = L + a. Here we place the test point at the origin of the system

E = k λ (-1 / x)

E = k λ (-1 /(L + a) + 1 /a)

E = k λ (L /a(L + a)

Let's change the density for its value

E = k (q / L) (L / a (L + a)

E = k q 1 /[a(L + a)]

E = 8.99 10⁹ 4.23 10⁻¹⁵ [1 /12 10⁻²(8.15 10⁻² + 12 10⁻²)]

E = 1,573 10⁻³ N/C

c) the direction of the field is directed to the bar, because it has a negative charge

d) now we change the distance a = 50 cm = 0.50 m

Bar

E = 8.99 10⁹ 4.23 10⁻¹⁵ ( 1 /0.5(0.0815 +0.5))

E = 1,308 10⁻⁴ N/C

Charge point

q = -4.23 10⁻¹⁵ C

E = k q / r²

E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

E = 1.521 10⁻⁴ N/C

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3 years ago

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Eduardwww [97]

Answer:

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3 years ago

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Planet X has three times the free-fall acceleration of Earth.

serious [3.7K]

Answer:

a) The ball goes one-third times higher on X

b) The ball goes three times higher on X.

Explanation:

As the initial velocity is the same than on Earth, but the free-fall acceleration is three times larger, this means that the only net force acting on the ball (gravity) will be three times larger, so it is clear that the ball will reach to a lower height, as it will slowed down more quickly.
Kinematically, as we know that the speed becomes zero when the ball reaches to the maximum height, we can use the following kinematic equation:

$v_{f} ^{2} - v_{o}^{2} = 2* \Delta h* g$

since vf = 0, solving for Δh, we have:

$\Delta h = h_{max} =\frac{v_{o} ^{2}}{2*g} (1)$

if v₀ₓ = v₀E, and gₓ = 3*gE, replacing in (1), we get:

Δhₓ = 1/3 * ΔhE

which confirms our intuitive reasoning.

Now, if the initial velocity is three times larger than the one on Earth, even the acceleration due to gravity is three times larger, we conclude that the ball will go higher than on Earth.
We can use the same kinematic equation as in (1) replacing Vox by 3*VoE, as follows:

$\Delta h = h_{max} =\frac{(3*v_{o}) ^{2}}{2*3*g} (2)$

Replacing the right side of (1) in (2), we get:

Δhx = 3* ΔhE

which confirms our intuitive reasoning also.

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3 years ago