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3 years ago

Which electromagnet is the strongest

Physics

2 answers:

Ahat [919]3 years ago

5 0

Answer: the second one

Explanation:

aleksandrvk [35]3 years ago

3 0

In the electromagnetic spectrum, we go and learn about rays going in a specific order. For example the least strong ER is Radiowave, and the strongest bring Gamma Rays.

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With an angular velocity of 1.7π rad/s, how much time is required to complete 18 complete revolutions?

sineoko [7]

The angle for each complete revolution is 2π radians (that's 360°).

So . . .

Time = (18 revs) x (2π radians / rev) x (1 sec / 1.7π radians)

Time = (18 · 2π · 1 / 1.7π) (rev ·radian · sec / rev · radian)

<em>Time = 21.18 seconds</em>

5 0

3 years ago

A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.00 m at constant speed. If the coeff

zysi [14]

Answer:

Work done is equal to 125.44 J

Explanation:

We have given mass of the student m = 80 kg

Distance moved d = 4 m

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of kinetic friction $\mu =0.4$

So normal force exerted by student $F=\mu mg=0.4\times 80\times 9.8=313.6N$

We know that work done is equal to multiplication of force and distance

So work done $W=Fd=313.6\times 0.4=125.44J$

7 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

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

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0

3 years ago

Two satellites are in circular orbits around a planet that has radius 9.00×106 m. One satellite has mass 68.0 kg, orbital radius

olasank [31]

Answer:

v₂ = 5131.42 m/s

Explanation:

given,

radius of the planet = r₁ = 9.00×10⁶ m

mass of the satellite = m₁ = 68 Kg

orbital radius = r₁ = 8 x 10⁷ m

orbital speed = v₁ = 4800 m/s

mass of second satellite = m₂ = 84.0 kg

orbital radius = r₂ = 7.00×10⁷ m

orbital speed of second satellite = v₂ = ?

using orbital speed of satellite

$v = \sqrt{\dfrac{GM}{r}}$

so,

$v \alpha \dfrac{1}{\sqrt{r}}$

now,

$\dfrac{v_1}{v_2} =\sqrt{\dfrac{r_2}{r_1}}$

$\dfrac{v_2}{4800} =\sqrt{\dfrac{8\times 10^7}{7 \times 10^7}}$

v₂ = 5131.42 m/s

The orbital speed of second satellite is equal to v₂ = 5131.42 m/s

7 0

3 years ago

A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of

lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = $20.6\times 10^3\ m$

$\theta$ = The angle subtended by moon = $9.6\times 10^{-3}\ rad$

Distance traveled is given by

$s=r\times\theta$

$\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}$

$\Rightarrow s=197.76\ m$

The distance traveled by the jet is 197.76 m

8 0

3 years ago