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Marina CMI [18]
3 years ago
15

An object is hanging by a string from the ceiling of an elevator. the elevator is moving upward with a constant speed. what is t

he magnitude of the tension in the string? an object is hanging by a string from the ceiling of an elevator. the elevator is moving upward with a constant speed. what is the magnitude of the tension in the string?
Physics
2 answers:
Anarel [89]3 years ago
6 0

Since the elevator is moving with a constant speed and not accelerating, the tension in the string is simply the normal, routine, everyday boring weight of the object.  Since the elevator is moving with a constant speed and not accelerating, the tension in the string is simply the normal, routine, everyday boring weight of the object.  

Molodets [167]3 years ago
6 0

Answer:

As elevator has no acceleration as it is moving upward with constant speed as states in the problems.Using equilibrium equation

∑Fy=0

T-W=0

T=W of object in the elevator.

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Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?​
IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity  = 3.8km/h

Unknown:

time  = ?

Solution:

Velocity is displacement divided by time.

  Velocity  = \frac{displacement}{time}  

      Displacement  = velocity x time

Input the parameters:

              8  = 3.8  x time

 Time  = \frac{8}{3.8}   = 2.1s

The time taken is 2.1s

6 0
3 years ago
Which is a scalar quantity displacement distance force acceleration​
cestrela7 [59]

A scalar is a quantity that is fully described by a magnitude only. It is described by just a single number. Some examples of scalar quantities include speed, volume, mass, temperature, power, energy, and time. A vector is a quantity that has both a magnitude and a direction.

I hope this helps you.

5 0
3 years ago
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
What type of galaxy is this?
stiks02 [169]

Answer:

Barred Spiral.

8 0
2 years ago
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va
Hitman42 [59]

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

5 0
3 years ago
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