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8_murik_8 [283]
2 years ago
11

The ant's aunt's aunt travels from the 15.00 cm mark to the 85.00 cm mark back to the 30.00 cm mark and then finally to the 30.0

0 cm mark. What is the ant's displacement?
Physics
1 answer:
navik [9.2K]2 years ago
5 0

Answer:

<em>15 cm</em>

<em></em>

Explanation:

Displacement is the shortest distance between the point of origin of a body in motion to the final stopping positing.

For this ant's aunt's aunt that travels from the 15.00 cm mark to the 85.00 cm mark back to the 30.00 cm mark and then finally to the 30.00 cm mark, the displacement from the starting position is..

displacement = stopping point - starting point

d = 30 cm - 15 cm = <em>15 cm</em>

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A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
Which of the following selections completes the following nuclear reaction?
Monica [59]

Answer:

n (a neutron)

Explanation:

For a chemical element:

- The lower subscript indicates the atomic number (the number of protons)

- The upper subscript indicates the mass number (the sum of protons and neutrons in the nucleus)

In the reaction described in the problem, we see that a gamma photon hits a nucleus of Calcium-40, which has

Z = 20 (20 protons)

A = 40 (40 protons+neutrons)

Which means that the number of neutrons is n =  A - Z = 40 - 20 = 20

After the reaction, we have a nucleus of Calcium-39, which has

Z = 20 (20 protons)

A = 39 (39 protons+neutrons)

Which means that the number of neutrons is n =  A - Z = 40 - 39 = 19

So, the nucleus has lost 1 neutron, which is the particle missing in the reaction.

3 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

6 0
2 years ago
Read 2 more answers
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