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Thepotemich [5.8K]
3 years ago
15

Santa doesn't want to push a 25.0 kg wooden box across a wooden floor with a uk of 0.20 at

Physics
1 answer:
scZoUnD [109]3 years ago
6 0

Answer:

49 N

Explanation:

In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,

F = ma

that the net force acting on the box, F, must be zero as well.

Here there are two forces acting on the box in the horizontal direction while it is moving:

- The force of push applied by the guy, F

- The frictional force, F_f

For an object moving on a flat surface, the frictional force is given by

F_f = \mu_k mg

where

\mu_k is the coefficient of friction

m is the mass of the box

g is the acceleration of gravity

So the equation of the forces becomes

F-\mu_k mg = 0

And substituting:

\mu_k = 0.20\\m = 25.0 kg\\g = 9.8 m/s^2

We find the force that must be applied by the guy:

F=\mu_k mg = (0.20)(25.0)(9.8)=49 N

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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0
3 years ago
PLS HELP<br><br> compare longitudinal and transverse waves
babymother [125]
<span>A transverse wave is one for which the direction of oscillation is perpendicular to the direction of propagation of the wave whereas, for longitudinalwaves oscillations are in the direction of propagation. Ripples in pond water move about the surface of water and they simultaneously move away from the point-0 too.</span><span>
Longitudinal waves include sound waves(vibrations in pressure, particle of displacement, and particle velocity propagated in an elastic medium) and seismic P-waves (created by earthquakes and explosions). In longitudinal waves, the displacement of the medium is parallel to the propagation of thewave.


</span>
6 0
3 years ago
Read 2 more answers
A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in met
Tamiku [17]

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

6 0
3 years ago
An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then
gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

a=\frac{dv}{dt}=1.76

So, the velocity can be obtained by integrating this expression:

v=1.76t

The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

At this time the velocity is: v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}

This velocity remains constant in the section 2, so for that section the movement equation is:

x=v*t\\t=\frac{x}{v}

The left distance is 89 meters, and the velocity is 6.2225\frac{m}{s}, so:

t=\frac{89}{6.2225}=14.303s

So, the total time is 14.303+3.5355s=17.838s

7 0
3 years ago
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal sp
gulaghasi [49]

Answer:

C

Explanation:

horizintal speed stays same

only vertical speed changes

so H speed will stay 30 m/s

6 0
3 years ago
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