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emmainna [20.7K]
2 years ago
11

What is the name

Physics
1 answer:
Molodets [167]2 years ago
3 0

Answer:

<h3>It's called Resistance! </h3>

Explanation:

If the ratio is constant over a wide range of voltages, the material is said to "ohmic" material.

Hope it helps!

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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
You lift a large bag of flour from the floor to a 2.5m high counter, doing 400J of work in 2 seconds. How much force did you app
oksian1 [2.3K]
<h2>The man have to apply force of 160 N</h2>

Explanation:

The work done to lift the bag of weight mg through height 2.5 m is 400 J

The work done can be found by relation  W = mg x h

Thus mg = \frac{W}{h} = \frac{400}{2.5} =  160 N

Therefore the man have to apply the force of 160 N

7 0
3 years ago
Read 2 more answers
2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t
serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0
2 years ago
what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
2 years ago
Read 2 more answers
A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule
saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

m_1v_1+m_2v_2=0

170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0

v_2=-0.17\ m/s

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0
3 years ago
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