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2 years ago

Which of these terms means an object's resistance to changing its motion?

Physics

1 answer:

Mila [183]2 years ago

8 0

Answer:

Inertia

F=ma

Action, reaction

All of the above

A heavy object requires more force to push than a lighter object.

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How much work is done if a force of 20N is used to move an object 6 metres? <br><br> pls help

Dmitriy789 [7]

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

8 0

2 years ago

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Urination is one of the key mechanisms for losing fluid from the human body.

serious [3.7K]

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3 years ago

Read 2 more answers

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co

ladessa [460]

Answer:

$\dot W_{in} = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_{u}\cdot T$

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$

$\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}$

$\rho = 1.2\,\frac{kg}{m^{3}}$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )$

$\dot m = 0.18\,\frac{kg}{s}$

The work input is:

$\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)$

$\dot W_{in} = 49.386\,kW$

5 0

2 years ago

How much energy does it take to boil water for pasta? For a one-pound box of pasta

alexandr1967 [171]

Answer:

a. 164°F

b. $91.\overline 1 \ ^{\circ} C$

c. $140.\overline 4$ kJ

Explanation:

The starting temperature of the water, T₁ = 48F

The temperature at which the water boils, T₂ = 212°F

a. The difference between the initial and the boiling water temperature, ΔT = T₂ - T₁

Therefore;

ΔT = 212°F - 48°F = 164°F

The temperature by which he temperature must be raised, ΔT = 164°F

b. 48°F = ((48 - 32)×5/9)°C = (80/9)°C = $8.\overline 8 \ ^{\circ} C$

212°F = ((212 - 32)×5/9)°C = 100°C

∴ ΔT = 100°C - $8.\overline 8 \ ^{\circ} C$ = $9.\overline 1 \ ^{\circ} C$

c. The heat capacity of the water = The heat required to increase four quartz of water by 1 °C = 15.8 kJ

∴ The heat required to raise four quartz of water by $9.\overline 1 \ ^{\circ} C$ , ΔQ = 15.8 kJ/°C × $9.\overline 1 \ ^{\circ} C$ = $140.\overline 4$ kJ.

5 0

2 years ago

2. Bank robbers have pushed a 1000 kg safe to a second story floor-to-ceiling window. They plan to break the window, then lower

zlopas [31]

Answer:

5.4 m/s

Explanation:

Given that

Mass of the safe, m1 = 1000 kg

Distance to lower the safe, d = 3 m

Mass of furniture, m2 = 500 kg.

Speed of the safe, v = ?

To get the final speed by the time that the safe hits the truck, we first find its acceleration.

The total mass of the system is M = 1000 + 500 kg = 1500 kg

One of the forces acting on the system is that of gravity, and it acts on the safe friction acting on the furniture. Using the formula, we have

= m1*g - mu*m2g

= 1000 * 9.81 - 0.5 * 500 * 9.81

= 7357.5 N

From this calculated weight, we find the acceleration.

Acceleration, a = F/m

Acceleration, a = 7357.5 / 1500

Acceleration, a = 4.905 m/s²

From the question, we know that the Initial speed = 0 m/s

So, employing the use of one of the equations of motion, we have

v² - u² = 2aS

v² - 0 = 2 * 4.905 * 3

v² = 29.43

v = √29.43

v = 5.4 m/s

5 0

2 years ago