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andrew-mc [135]

3 years ago

9

Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An

engineer tests a cell phone and finds that the batteries supply 10,000 J of energy to make 5500 J of output energy in the form of sound and light for the screen. How efficient is the phone?

1 answer:

Svetradugi [14.3K]3 years ago

5 0

efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

=55%

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Determine the gravitational force that you exert on another person 1.50 m away. assume that you and the other person are point m

Aneli [31]

Given: Universal Gravitational constant = G = 6.67 x 10⁻¹¹ N m²/Kg²

Mass₁ = 70 Kg; Mass₂ = 70 Kg Radius r = 1.5 m; Force F = ?

Formula: F = Gm₁m₂/r²

F = (6.67 x 10⁻¹¹ N m²/Kg²)(70 Kg)(70 Kg)/(1.5 m)²

F = 1.45 X 10⁻⁷ N

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3 years ago

What is a cluster?

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A cluster is D.) Portion of space containing many galaxies.

Hope this helps. Chao.

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3 years ago

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a

xxTIMURxx [149]

Answers:

a) $F_{g}=735 N$ and $n=732.47 N$ , hence $F_{g} > n$

b) $n_{poles}=735 N$ $n_{equator}=732.47 N$

Explanation:

a) At the equator, both the <u>centripetal force</u> $F_{c}$ and the <u>gravitational force</u> $F_{g}$ (also called true weight) are directed "downward", while the <u>normal force</u> $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation:

$n_{equator}-F_{g}=-F_{c}$ (1)

Where:

$F_{g}=m g$ being $m=75 kg$ the mass and $g=9.8 m/s^{2}$ the acceleration due gravity

$F_{c}=m a_{c}$ being $a_{c}=0.0337 m/s^{2}$ the centripetal acceleration at the equator

According to this (1) is rewritten as:

$n_{equator}-mg=-m a_{c}$ (2)

Isolating $n_{equator}$ :

$n_{equator}=-m a_{c} + mg$ (3)

$n_{equator}=m(-a_{c}+g)$ (4)

$n_{equator}=75 kg (-0.0337 m/s^{2}+9.8 m/s^{2})$ (5)

$n_{equator}=732.47 N$ (6) This is the apparent weight at the equator

The true weight is given by $F_{g}=m g=75 kg (9.8 m/s^{2})$

Hence: $F_{g}=735 N$ (7)

As we can see $F_{g} > n_{equator}$

b) Now we have to calculate the apparent weight at the poles $n_{poles}$ :

$n_{poles}-F_{g}=-F_{c-poles}$ (8)

Since $F_{c-poles}=0$ (8) is rewritten as:

$n_{poles}=F_{g}$ (9)

$n_{poles}=m g$ (10)

$n_{poles}=(75 kg)(9.8 m/s^{2})=735 N$ (11)

So, the apparent weight of the person at the poles is 735 N and at the equator is 732.47 N

5 0

3 years ago

Crossed electric and magnetic fields are established over a certain region. The magnetic field is 0.669 T vertically downward. T

ch4aika [34]

Answer:

The speed is $v = 3.946 *10^6 \ m/s$

Explanation:

From the question we are told that

The magnetic field is $B = 0.669\ T$

The electric field is $E = 2.64 *10^{6} \ V/m$

Generally the speed of the electron is mathematically represented as

$v = \frac{E}{B}$

Substituting values

$v = \frac{2.64 *10^6}{0.669}$

$v = 3.946 *10^6 \ m/s$

8 0

4 years ago

Express 9.39 x10^9 seconds I’m terms of days

iogann1982 [59]

Converting seconds into days is very simple

We have 60 Second in 1 minutes and 60 minutes in 1 hour and 24 hours in one day therefore
Days = 9,390,000,000/60x60x24

Days = 108680.55

6 0

3 years ago