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solniwko [45]
3 years ago
13

How much larger is the acceleration with two people pushing on one side of the cart compared to only one person pushing the cart

?
Physics
1 answer:
il63 [147K]3 years ago
6 0
As we said to the two people name A and the single person nameB.as every person make an acceleration 60ms2.so group A has more acceleration then group B. because in A there are two person so we add 60+60=120ms2. so group A can produce 120ms2 while groupB can produce 60ms2. i hope it will help you.

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True or False?
WINSTONCH [101]

Answer:

true is the answer of the question

5 0
2 years ago
Read 2 more answers
You have a length of tubing that is closed at one end. You cut the tubing into two pieces of unequal length, giving you a tube o
umka21 [38]

Answer:

The funda mental frequency of the original tube is 182Hz.

Explanation:

See the attachment for the calculation steps.

In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.

For closed tubes

f = nv/4L (n = 1, 3, 5,...n)

f = nv/2L (n = 1, 2, 3,...n)

The details of calculation can be found below in the attachment.

4 0
3 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
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4 0
2 years ago
A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0
3 years ago
Your cell phone typically consumes about 390 mW of power when you text a friend. If the phone is operated using a lithium-ion ba
yulyashka [42]

Answer:

I = 0.11 A

Explanation:

  • In an electric circuit, the power delivered to a load, is just the product of the potential difference between the load terminals, times the current flowing through it, as follows:

       P = V*I

  • In this case, the power is the one consumed by the cell phone = 390 mW, and the voltage the one produced by the internal energy of the battery, 3.5 V, neglecting the voltage loss at the internal resistance of the battery.
  • So, we can solve the above equation for  the current I, as follows:

        I = \frac{P}{V} = \frac{0.39W}{3.5V}  = 0.11 A

  • The current flowing through the cell-phone circuitry is 0.11 A.
4 0
2 years ago
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