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Oduvanchick [21]

2 years ago

13

While John is traveling along a straight interstate

1 answer:

lora16 [44]2 years ago

7 0

Answi dont know

Explanation:

i dont know

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3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

8 months ago

SI unit for electrical current

Mashcka [7]

I believe it is called an ampere.

5 0

2 years ago

Which of the following is one way that an electric motor's rotation speed can be controlled?

malfutka [58]

Depending on which type of motor you're talking about, but the first 3 are true. A stronger magnetic field in a DC motor will slow it down but increase its torque.
The amount of current in the motor will control the magnetic fields and therefore affect the speed (and torque). In an induction motor, the rotational speed is given by $n= \frac{120f}{p}$ where f is the line frequency and p is the number of poles. Thus fewer poles makes it go faster.

5 0

2 years ago

Read 2 more answers

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

2 years ago

A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati

sp2606 [1]

Answer:

a) $T = 2.9*10^{-5} N-m$

b) north edge will rise up

Explanation:

torque on the coil is given as

$T = NIABsin\theta$

where N is number of loop = 9 loops

i is current = 7.80 A

-B -earth magnetic field = $5.00*10^{-5} T$

A- area of circular coil

$A = \frac{\pi d^{2}}{4}$

$A =\frac{\pi .14^{2}}{4}$

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

$T = 9*7.8*0.015*5*10^{-5} sin{90-56}$

$T = 2.9*10^{-5} N-m$

b) north edge will rise up

3 0

2 years ago

Read 2 more answers