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3 years ago

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A man pushes a cart with force of 120-N at an angle of 50 degrees with the ground. If he walks a horizontal displacement of 20-m

going to cashier who’s much work is done!

1 answer:

Travka [436]3 years ago

7 0

Answer:

1543 J

Explanation:

The work done by a force when moving an object is given by the equation:

$W=Fd cos \theta$

where:

F is the magnitude of the force applied on the object

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the cart pushed in this problem, we have:

$F=120 N$ is the force applied by the man

$d=20 m$ is the displacement of the cart

$\theta=50^{\circ}$ is the angle between the directions of the force and of the displacement

Solving for W, we find the work:

$W=(120)(20)(cos 50^{\circ})=1543 J$

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A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago

This is PLZ HELP MY MOM GONNA GROUND ME!!!! Which step is part of a scientific investigation?

Novosadov [1.4K]

Answer:

evalating the soltion

Explanation:

4 0

4 years ago

Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

The single interface equation is $\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}$

Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

The image distance is then $y^i=-\frac{18}{4}in =-4.5in$

Therefore, the coin falls $1.5in$ in front of the target

6 0

3 years ago

While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38° with the vertical. While driving

Natali [406]

Answer: $21.33^{\circ}$

Explanation:

Given

velocity of driver $v_1$ =25 m/s w.r.t ground towards north

driver observes that rain is making an angle of $38^{\circ}$ with vertical

While returning $v_2$ =25 m/s w.r.t. ground towards south

suppose $u_1$ =velocity of rain drop relative car while car is going towards north

$u_2$ =velocity of rain drop relative car while car is going towards south

z=vector sum of $u_1 & v_1$

Now from graph

$\tan 38=\frac{v_1+v_2}{u_2}$

$u_2=\frac{25+25}{\tan 38}=64 m/s$

$z=\vec{u_2}+\vec{v_2}$

therefore magnitude of z is given by

$|z|=\sqrt{u_2^2+v_2^2}$

$|z|=\sqrt{64^2+25^2}$

$|z|=68.70 m/s$

$\tan A=\frac{v_2}{u_2}$

$\tan A=\frac{25}{64}=0.3906$

$A=21.33^{\circ}$

Thus rain drops make an angle of $21.33^{\circ}$ w.r.t to ground

6 0

3 years ago

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0

3 years ago