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2 years ago

13

Which of the following boxes would have the strongest gravitational pull on a third object positioned at an equal distance from

both boxes?
Box A: 100 g
Box B: 10,000 g
A. box b because it has more mass
B. box a because it has less mass
C. box a because it is bigger
D. box b because it is smaller

1 answer:

trasher [3.6K]2 years ago

3 0

<span>A. box b because it has more mass </span>

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An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0

2 years ago

A flashlight converts?

sammy [17]

Answer:

Energy conversion in flashlight.

Explanation:

When the switch of a flashlight is on, then the batteries inside them will be connected in a circuit and chemical energy from the batteries will be converted into electrical energy and then the electrical energy will be converted into light energy.

Light energy is also called electromagnetic energy.

Hence, among the given options the correct one is the first option that is:

1. Electrical energy into electromagnetic energy is the answer.

4 0

1 year ago

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0

2 years ago

The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is:

$F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}$

$F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}$

$F = 23437500\,N$

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0

2 years ago

Can someone help me with universal gravation

Nesterboy [21]

States that particles are attracts with every other particle. wich force is directily proportional product of two masses and inversely proportional to the distance between the centers.

6 0

2 years ago