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kondaur [170]
3 years ago
13

A 55.0 kg student feels a

Physics
1 answer:
natta225 [31]3 years ago
5 0

Explanation:

mass of student(M1)=55

mass of book(M2)=?

distance(d)=0.435m

gravitational constant (G)=6.67×10^-11

force(F)=2.68×10^-8

Now,

F=GM1M2/d^2.

then do the numericals .It's taking long time so I escaped Sorry

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A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.90
kiruha [24]

Answer:

1 m/s² down

Explanation:

"Calculate the acceleration of the elevator, and find the direction of acceleration."

Sum of forces in the +y direction:

∑F = ma

N − mg = ma

0.9 mg − mg = ma

-0.1 mg = ma

a = -0.1 g

a = -1 m/s²

6 0
3 years ago
If a data point is way off the trend line, what will not resolve the problem?
Elden [556K]

Going out to a movie will not resolve the problem.

4 0
3 years ago
A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c
DedPeter [7]

Answer:

The induced current in the coil at the time 2 s is 0.00263 A

Explanation:

The equation for induced emf is equal to:

\epsilon =-\frac{d}{dt} (BAcos\theta )

Where

B = magnetic field

A = area

θ = angle

\epsilon =-Acos0\frac{d}{dt} B\\\epsilon =-(\pi r^{2} )\frac{d}{dt} (0.5e^{-0.2t} )\\\epsilon =0.1 (\pi r^{2} )e^{-0.2t}

For t = 2 s

\epsilon =0.1*\pi *0.05^{2} *e^{-0.2*2} =5.26x10^{-4} V

The induced current is:

I=\frac{\epsilo }{R} =\frac{5.26x10^{-4} }{0.2} =0.00263A

3 0
3 years ago
Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce
Stells [14]

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

\omega=\frac{v}{r}

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}

the maximum relative velocity is 0.04m/s

hope this helps!!

6 0
3 years ago
Read 2 more answers
A particle of mass 10 g and charge 80 mC moves through a uniform magnetic field,in a region where the free-fall acceleration is .
Alex777 [14]

Answer:

B=1.223\frac{T\cdot m}{s}\frac{1}{v}

Explanation:

According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:

F_m=F_g

Here F_m=qvBsin\theta is the magnetic force and F_g=mg is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so \theta=90^\circ. Replacing and solving for B:

qvBsin(90^\circ)=mg\\B=\frac{mg}{qv}\\B=\frac{mg}{q}\frac{1}{v}\\B=\frac{(10*10^{-3}kg)(9.8\frac{m}{s^2})}{80*10^{-3}C}\frac{1}{v}\\B=1.223\frac{T\cdot m}{s}\frac{1}{v}

4 0
3 years ago
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