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1 day ago

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There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

ter. If the smaller paintball holds 28 cm3 of paint, how much does the larger one hold?

1 answer:

Ostrovityanka [942]1 day ago

5 0

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Diameter of the smaller paintball = 5 cm

d₂ = Diameter of the larger paintball = 9 cm

V₂ = Volume of the larger paintball

Volume of the smaller paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Likewise

$V_2=\frac{4}{24}\pi d_2^3$

By dividing the previous two equations, we derive

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one accommodates 163.296 cm³ of paint

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Answer:

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Explanation:

The mass of each ball is $m= 10 g = 0.01 \ kg$

The negative charge of each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Since the lower ball is prevented from moving, it indicates that the net force acting on it equals zero.

This means that the gravitational force exerted on the lower ball balances the electrostatic force, represented as:

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A household refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h.

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Answer:

1454.54 kJ/h or 0.404 kW

Explanation:

Provided:

The heat being removed by the refrigerator is 800 kJ/h.

The refrigerator's coefficient of performance (COP) is 2.2.

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Next, the power consumed by the refrigerator during its operation is calculated as:

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= $\frac{3200 kJ/h}{2.2}$ .

= 1454.54 kJ/h

Thus, the fridge uses 1454.54 kJ/h

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Pilobolus is a genus of fungi commonly found on dung and known for launching its spores a large distance for a sporangiophore on

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1) The work performed on the spores is $2.45\cdot 10^{-7}J$

2) The spores land 0.32 m away

Explanation:

1)

According to the work-energy principle, the work done on the spores equals their change in kinetic energy:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is work done

m represents the spore's mass

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u is the initial velocity

Given:

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initial velocity u = 0 (starting from rest)

final velocity v = 7.0 m/s

Calculating the work:

$W=\frac{1}{2}(10^{-8})(7.0)^2-0=2.45\cdot 10^{-7}J$

2)

The spores follow projectile motion, moving along a parabolic trajectory made of two motions:

- Constant speed horizontally

- Vertically accelerated due to gravity

Considering vertical motion first, using kinematics:

$s=ut+\frac{1}{2}at^2$

where

s = 0.01 m (shooting height)

u = 0 (no initial vertical velocity, horizontal ejection)

t = time of flight

g = acceleration due to gravity $a=g=10 m/s^2$

Solving for t:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.01)}{10}}=0.045 s$

Now for horizontal displacement, when velocity is constant:

$d=v_x t$

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time t = 0.045 s

Calculating distance d:

$d=(7.0)(0.045)=0.32 m$

Thus, the spores land 0.32 meters away.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

Learn more about projectile motion:

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A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. R

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Answer:

Dₓ = -155 sin 23° i + 0 j

Explanation:

The accompanying diagram illustrates the vector.

According to the diagram,

The vector D has an x-component (or horizontal component) expressed as -D sinθ i. Specifically,

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Where;

D = magnitude of D = 155m

θ = angle of D = 23°

Consequently;

Dₓ = -155 sin 23° i

Since Dₓ signifies the x component, its unit vector, j component holds a value of 0.

<pthus d="" can="" be="" formulated="" in="" terms="" of="" and="" the="" unit="" vectors="" i="" j="" as="" follows="">

Dₓ = -155 sin 23° i + 0 j

</pthus>

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