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zalisa [80]
3 years ago
5

Some tropical butterflies can reach speeds of up to 11 m/s. Suppose a butterfly flies at a speed of 6.0m/s while another flying

insect some distance ahead flies in the same direction with a constant speed. The butterfly then increases its speed at a constant rate of 1.4m/s2 and catches up to the other insect 3.0s later. How far the butterfly travel during the race
Physics
1 answer:
ladessa [460]3 years ago
7 0

Answer:

24.3 m

Explanation:

Using the equation of motion

S=ut+0.5at^{2} where s is the distance, u is initial velocity, t is time and a is acceleration

Substituting u for 6 m/s, t for 3 s, a for 1.4 m/s2 we obtain

S=6*3+(0.5*1.4*3^{2}=18+6.3=24.3 m


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n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend
Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

p_f=p_i\\mv_1+Mv_2=(m+M)v

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m

hence, the heigth is 68cm

4 0
3 years ago
g When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity
GREYUIT [131]

Answer:

This is because motion is intended to occur but at zero acceleration. It means at a constant velocity, henceFor that to happen the pulling force F must exactly equal the frictional force Fk .

7 0
3 years ago
A wire of Nichrome (a nickel– chromium– iron alloy commonly used in heating elements) is 1.0 m long and 1.0 mm² in cross-se
Lyrx [107]

Answer:

The conductivity of Nichrome is 2\times 10^6\ S/m.

Explanation:

Given:

Potential difference (V) = 2.0 V

Current flowing (I) = 4.0 A

Length of wire (L) = 1.0 m

Area of cross section of wire (A) = 1.0 mm² = 1 × 10⁻⁶ m² [1 mm² = 10⁻⁶ m²]

We know, from Ohm's law, that the ratio of voltage and current is always a constant and equal to the resistance of the resistor. Therefore, the resistance of the nichrome wire is given as:

R=\frac{V}{A}=\frac{2.0}{4.0}=0.5\ \Omega

Now, resistance of the nichrome wire in terms of its resistivity, length and area of cross section is given as:

R=\rho\frac{L}{A}

Where, \rho\to resistivity\ of\ Nichrome

Now, plug in all the values given and solve for \rho. This gives,

0.5\ \Omega=\rho\frac{1.0\ m}{1\times 10^{-6}\ m^2}\\\\\rho=\frac{0.5\times 1\times 10^{-6}}{1.0}=0.5\times 10^{-6}\ \Omega-m

Now, conductivity of a material is the reciprocal of its resistivity. Therefore, the conductivity of Nichrome is given as:

\sigma=\frac{1}{\rho}=\frac{1}{0.5\times 10^{-6}}=2\times 10^6\ S/m

Conductivity is measured in Siemens per meter (S/m)

Therefore, the conductivity of Nichrome is 2\times 10^6\ S/m.

7 0
3 years ago
A wall with a surface area of 10 m2 is 2.5 cm thick. The inner surface temperature of the wall is 4150C, but the outer surface i
klio [65]

Answer:

650.65 K or 377.5°C

Explanation:

Area = A = 10 m²

Thickness of wall = L = 2.5 cm = 2.5×10⁻² m

Inner surface temperature of wall = T_i = 415°C = 688.15 K

Outer surface temperature of wall = T_o

Heat loss through the wall = 3 kW = 3×10³ W

Thermal conductivity of wall = k = 0.2 W/m K

Assumptions made here as follows

  1. There is not heat generation in the wall itself
  2. The heat conduction is one dimensional
  3. Heat flow follows steady state
  4. The material has same properties in all directions i.e., it is homogeneous.

Considering the above assumptions we use the following formula

Q=\frac {T_i-T_o}{\frac{L}{kA}}\\\Rightarrow T_o=T_i-\frac {QL}{kA}\\\Rightarrow T_o=688.15-\frac {3\times 10^{3}\times 2.5\times 10^{-2}}{0.2\times 10}\\\Rightarrow T_o=650.65~K

∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C

5 0
4 years ago
Which equation shows the beta decay of a transuranium element?
kicyunya [14]

239        239     0

Np    →    Pu   +   e

93          94        -1  is the equation shows the beta decay of a transuranium element.

<h3>What is a Transuranium element?</h3>

These are elements whose atomic number is greater than that of uranium which is 92.

Neptunio has an atomic number of 93 which is why option D was chosen as the most appropriate choice.

Read more about beta decay here brainly.com/question/12448836

#SPJ1

5 0
2 years ago
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