A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What

Which of the following correctly arranges the planets by the size of their diameter in increasing order?

valentinak56 [21]

Increasing Order..
So, it's from the little one into the biggest one..

My Answer is
C: Mercury < Earth < Saturn < Jupiter

What Shape of The Earth ?
Is it Round or Flat ?
Subjective Answer..

7 0

3 years ago

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A 25 kg rock resting on the bottom of a lake must be moved from the paths of boats. The rock has a density of 2350 kg/m^3. What

algol13

Answer:

The force needed is the weight of the rock minus the buoyant force.

Explanation:

7 0

3 years ago

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).

sammy [17]

Answer:

(a) $emf_L=-LI_{max}\omega cos(\omega t)$

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

$I(t)=I_{max}sin(\omega t)$ (1)

(a) The induced emf is given by the following formula

$emf_L=-L\frac{dI}{dt}$ (2)

You derivative the expression (1) in the expression (2):

$emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)$

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

$emf_L=-\omega LI_{max}$

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0

2 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

ladessa [460]

A, it's more about the force with leverage.

3 0

3 years ago

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