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3 years ago

Which one of the following statements about attitudes is correct?

1 answer:

ozzi3 years ago

7 0

B is a correct answer. Positive attitudes can help us learn quicker. Hope it helped you, and have a great day.

-Charlie

You might be interested in

A child pulls on a wagon with a force of 75 N. If the wagon moves a total of 42 m in 3.1 min, what is the average power delivere

exis [7]

Answer:

16.96 W

Explanation:

Power: This can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).

From the question,

P = (F×d)/t....................... Equation 1

Where P = power, F = force, d = distance, t = time.

Given: F = 75 N, d = 42 m, t = 3.1 min = 3.1×60 = 186 s

Substitute these values into equation 1

P = (75×42)/186

P = 16.94 W

Hence the average power delivered by the child = 16.96 W

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:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

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

3 years ago

the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is

garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0

3 years ago

If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:

Orlov [11]

I could be wrong but I believe it’s 1/2

6 0

3 years ago

A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti

melisa1 [442]

1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons

4 0

3 years ago