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Svetradugi [14.3K]

2 years ago

8

What is the approximate efficiency of the engine?

2 answers:

Elodia [21]2 years ago

8 0

Answer:91% I think

Explanation:

MArishka [77]2 years ago

4 0

Answer:

B) 91%

Explanation:

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6. A lumberjack is standing on a log floating on a lake. She starts from rest, then runs along the log to the end, when she jump

scoray [572]

Answer:

a) -3.267 m/s

b) 2.227 m/s

Explanation:

As per the conservation of momentum

m1v1 + m2v2=0

m1= mass of log

m2 = mass of lumber jack

v1 = velocity of log

v2 = velocity of lumber jack

a) Velocity of first log

$-\frac{70*7}{150} = -3.267$ m/s

b) m1v1 + m2v2 = m3v3

Velocity of log

= $\frac{70*7}{150+70} \\2.227$

4 0

2 years ago

All atoms of the element potassium have 19 protons. One of the most stable types of potassium atoms has the mass number 39. How

nydimaria [60]

For atoms in the periodic table, the given mass number is the sum of the number of the protons (also called the atomic number) and the number of the neutrons inside its nucleus. Mathematically,
mass number = atomic number + number of neutrons
Substituting,
39 = 19 + n
n = 39 - 19 = 20
Therefore, the answer is not found in the choices.

7 0

2 years ago

Read 2 more answers

A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction

Nuetrik [128]

Answer:

brainly.com/question/5055678

Explanation:

4 0

2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

2 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago