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ella [17]
3 years ago
8

A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting attime t = 1.00 s. In the SI sy

stem, A has the numerical value 4.50 and B has the numerical value8.75. What impulse does this force impart to the object?
Physics
2 answers:
PSYCHO15rus [73]3 years ago
7 0

Answer:

3.82 Ns

Explanation:

Time varying horizontal Force is given as

F(t) = A t⁴ + B t²

F(t) = 4.50 t⁴ + 8.75 t²

Impulse imparted is given as

I = \int_{0}^{t}Fdt

I = \int_{0}^{1}Fdt

I = \int_{0}^{1}(4.50 t^{4} + 8.75 t^{2})dt

I = ((0.9) (1)^{5} + (2.92) (1)^{3})

I = 3.82 N-s

zlopas [31]3 years ago
7 0

Answer:

impulse is 12.8614 kg

Explanation:

Given data

F(t) = At4 + Bt2

time t = 1.00 s

A = 4.50

B = 8.75

to find out

What impulse does this force impart to the object

solution

we know impulse is the change in momentum so we can right this as that

impulse I  = F i.e

dI = F(t) dt

we integrate it with limit 1 to 1.5

I = \int_{1}^{1.5} At^4 + Bt^2

I = (At^5 / 5)^{1.5} _1 + (Bt^3 / 3)^{1.5} _1

put the value A and B

I = (4.50(1.5)^5 / 5) - (4.50(1)^5 / 5)  + (8.75(1.5)^3 / 3) -  (8.75(1)^3 / 3)

I = (4.50(1.5)^5 / 5) - (4.50(1)^5 / 5)  + (8.75(1.5)^3 / 3) -  (8.75(1)^3 / 3)

I = 6.8343 - 0.9 + 9.8437 - 2.9166

impulse is 12.8614 kg

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Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the
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The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire. 

The magnetic field generated by wire 1 is:
B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}=  4 \cdot 10^{-5}T

The magnetic field generated by wire 2 is:
B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T

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3 years ago
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

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