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tia_tia [17]
4 years ago
10

A light bulb that is connected to a 120 V source drawers 1.7 A. What is the resistance of the light bulb?

Physics
1 answer:
Tamiku [17]4 years ago
8 0
Ohm’s law says V=IR so if V is 120 and I is 1.7 divide voltage by current and your left with roughly 71
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Your body temperature begins to rise during strenuous exercise in hot weather. Specialized nerve cells on the skin detect an inc
Orlov [11]

Answer:

receptors

Explanation:

In this case, your skin and brain act as receptors in maintaining homeostasis. These receptors are monitors in different parts of an organism's body that are there to detect specific changes that occur and activate very specific responses in order to counteract these changes and maintain a very stable system. This is exactly, what your skin and brain are doing in this scenario, they detect the change in temperature and counteract its effects through sweat which ultimately causes the opposite effect and maintains equilibrium in your body.

8 0
4 years ago
A person who is 180 cm tall and weighs 750 newtons is driving a car at a speed of 90 kilometers per hour over a distance of 80 k
satela [25.4K]

Answer:

5.90551 ft

168.6151 lbf

55.92354 mph

49.70981 mi

0.07491 lb/ft³

Explanation:

Convert cm to ft

1\ cm=\dfrac{1}{30.48}\ ft

180\ cm=180\times \dfrac{1}{30.48}\ ft=5.90551\ ft

180 cm = 5.90551 ft

Convert N to lbf

1\ N=\dfrac{1}{4.448}\ lbf

750\ N=750\times \dfrac{1}{4.448}\ lbf=168.6151\ lbf

750 N = 168.6151 lbf

Convert kmph to mph

1\ kmph=\dfrac{1}{1.60934}\ mph

90\ kmph=90\times \dfrac{1}{1.60934}=55.92354\ mph

90 kmph = 55.92354 mph

Convert km to mi

1\ km=\dfrac{1}{1.60934}\ mi

80\ km=80\times \dfrac{1}{1.60934}=49.70981\ mi

80 km = 49.70981 mi

Convert °C to °F

^{\circ}C=(^{\circ}C\times \dfrac{9}{5})+32

30^{\circ}C=(30^{\circ}C\times \dfrac{9}{5})+32=86^{\circ}F

30°C = 86°F

Convert kg/m³ to lb/ft³

1\ kg/m^3=\dfrac{2.20462}{3.28084^3}\ lb/ft^3

1.2\ kg/m^3=1.2\times \dfrac{2.20462}{3.28084^3}=0.07491\ lb/ft^3

1.2 kg/m³ = 0.07491 lb/ft³

4 0
3 years ago
A pressure is applied to 2L of water. The volume is observed to decrease to 18.7 L. Calculate the applied pressure.
lakkis [162]

Answer:

Applied pressure is 1.08 10⁵ Pa

Explanation:

This exercise is a direct application of Boyle's law, which is the application of the state equation for the case of constant temperature.

PV = nR T

If T is constant, we write the expression for any two points

Po Vo = p1V1

From the statement the initial pressure is the atmospheric pressure 1.01 10⁵ Pa, so we clear and calculate

1 Pa = 1 N / m2

P1 = Po Vo / V1

P1 = 1.01 10⁵ 20/18.7

P1 = 1.08 10⁵ Pa

4 0
3 years ago
In which of the following situations is the most work done on the object?
aivan3 [116]
I believe D is ur answer..................
6 0
3 years ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
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