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3 years ago

8

A circuit consists of resistors 170 Ω, 240 Ω, and 65 Ω connected in series with a 12-V battery. What is the voltage across resis

tor 240 Ω?

1 answer:

Lyrx [107]3 years ago

5 0

Answer:

khgfdsdfghjhnbvcxzxcvbn

Explanation:

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PLS HELP WITH THIS FOR BRAINLIST IF ITS RIGHT

Zolol [24]

Answer:

the first one is Primary

the second one I think it's Mature but I don't know

5 0

3 years ago

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp

IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0

3 years ago

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Total number of oscillation per unit time is called frequency of oscillation. Here, $f=\dfrac{31}{15}=2.06\ Hz$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{15}{4\pi^2\times 2.06^2}$

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

$v_{max}=A\times \omega$

$v_{max}=A\times 2\pi f$

$v_{max}=0.075\times 2\pi \times 2.06$

$v_{max}=0.970\ m/s$

$v_{max}=97\ cm/s$

Hence, this is the required solution.

5 0

3 years ago

Technician A says mechanical efficiency is a comparison between brake horsepower and indicated horsepower. Technician B says vol

Reil [10]

Answer:Both are correct

Explanation:

Both are correct because

Mechanical efficiency is the dimensionless term which is the ratio of brake horsepower to the Indicated horse Power

Where brake power is the Power obtained at the crankshaft and

Indicated horsepower is the power obtained in the combustion chamber and this power is the loss in the form of friction.

Volumetric efficiency is the ratio of actual fuel intake to the maximum air fuel that could be taken.

8 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first cart $m_1=6 kg$

initial Velocity $u_1=3 m/s$

mass of second cart $m_2=3 kg$

$u_2=0 m/s$

In the absence of External Force we can conserve momentum

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\frac{6\times 3+3\times 0}{6+3}$

$v=2 m/s$

Final kinetic Energy of two masses

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2$

$K.E._2=18 J$

Initial Kinetic Energy

$K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$

$K.E._1=\frac{1}{2}6\times 3^2+0$

$K.E._1=27 J$

$ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5$

5 0

3 years ago