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labwork [276]
3 years ago
15

Suppose that 1000 pollen grains slam into the abdomen of a 5.0 g bee that is hovering just above the flower. the grain's mass is

1.0×10−7g. if the collision is inelastic, what is the bee's speed immediately afterward?

Physics
2 answers:
BaLLatris [955]3 years ago
7 0
<span>m1 = 5g = 0.005kg v1 = 0 m2 = 1.0 x 10^-7g = 1.0x10^-10kg v2 = 7.5 v3 = ? looks like you did the formula correctly m1v1 + m2v2 = (m1+m2)v3 0 + m2v2 = (m1+m2)v3 m2v2 / (m1+m2) = v3 (1.0x10^-10)(7.5) / (0.005 + 1.0x10^-10) = v3 1.5x10^-7 = v3 However, it said the bee gets hit by 1000 pollen grains, so I would guess you have to multiply 1.0 x 10^-10 by 1000 first, unless the total mass of the 1000 grains really is 1.0 x 10^-10.</span>
son4ous [18]3 years ago
6 0

The bee's speed immediately afterward is about 1.5 × 10⁻⁴ m/s

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Complete Question:</u>

Suppose that 1000 pollen grains slam with speed of 7.5 m/s into the abdomen of a 5.0 g bee that is hovering just above the flower. The grain's mass is 1.0×10⁻⁷ g. If the collision is inelastic, what is the bee's speed immediately afterward?

<u>Given:</u>

mass of pollen grains = m₁ = 1.0 × 10⁻⁷ g

number of pollen grains = N = 1000

mass of bee = m₂ = 5.0 g

initial speed of pollen grains = u₁ = 7.5 m/s

initial speed of bee = u₂ = 0 m/s

<u>Asked:</u>

final speed of bee = v = ?

<u>Solution:</u>

We will use law of conservation of momentum to solve this problem as follows:

Nm_1u_1 + m_2u_2 = Nm_1v_1 + m_2v_2

Nm_1u_1 + m_2u_2 = Nm_1v + m_2v

Nm_1u_1 + m_2u_2 = (Nm_1+m_2)v → <em>inelastic v₁ = v₂</em>

1000(1.0 \times 10^{-7})(7.5) + 5.0(0) = (1000(1.0 \times 10^{-7}) + 5.0)v

7.5 \times 10^{-4} + 0 = 5.0001v

v = 7.5 \times 10^{-4} \div 5.0001

v \approx 1.5 \times 10^{-4} \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
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Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
3 years ago
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