All categories

3 years ago

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

You might be interested in

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