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gregori [183]
3 years ago
7

Cassandra is asked to identify a solution in her laboratory classroom as acidic, basic, or neutral. She is given litmus

Physics
2 answers:
DerKrebs [107]3 years ago
7 0

Answer:

The blue litmus paper will turn blue

the solution is basic

Explanation:

avanturin [10]3 years ago
5 0

Answer:

Blue litmus paper remains blue

Solution is basic

Explanation:

Blue litmus paper turns red when a solution is acidic while the red litmus paper turns blue when a solution is alkaline.  In this case, Cassandra observed that red litmus paper turned blue, a clear indicator that the solution is alkaline. Similarly, when using blue litmus paper, the color remains blue for the same solution since it's basic.

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A vertical scale on a spring balance reads from 0 to 250 N. The scale has a length of 15.0 cm from the 0 to 250 N reading. A fis
lesantik [10]

Answer:

Mass of fish=6.252 kg

Explanation:

Force F=0 N to 250 N

Length x=15.0 cm=0.15 m

Frequency f=2.60 Hz

To find

Mass of fish m

Solution

First we need to find spring constant

k=Force/distance\\k=F/x\\k=250/0.15\\k=1666.7 N/m

As we know that the time period is given as

T=2\pi\sqrt{\frac{m}{k} }\\  And\\f_{Frequency} =1/T\\So\\1/f=2\pi\sqrt{\frac{m}{k} }\\1/f^{2}=4\pi^{2}\frac{m}{k}\\   k/f^{2}=4\pi^{2}m\\m=\frac{k}{4\pi^{2}f^{2} } \\m=\frac{1666.7}{4\pi^{2} (2.60)^{2} }\\ m=6.252kg

4 0
3 years ago
If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
  • force to move the object: F

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
Which botany phenomenon is primarily based on an understanding of physics
erik [133]
The botany phenomenon that is primarily based on an understanding of physics is t<span>he way plant pollen is carried by wind to another plant. It has to do with motion and acceleration, which are all the things that physics studies. </span>
3 0
4 years ago
Read 2 more answers
A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th
Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = .250\times 10^{- 3}\ m

Now,

Applying the equilibrium condition at the left end for torque:

T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}

T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}

T_{C} = 341.357\ Nm

The weight of the wire balances the tension in both the wires collectively:

W = T_{Al} + T_{C}

531 = T_{Al} + 341.357

T_{Al} = 189.643\ Nm

Now,

The fundamental frequency is given by:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

\mu = A\rho = \pi R^{2}\rho

(a) For the fundamental frequency of Aluminium:

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}

where

\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz

(b)  For the fundamental frequency of Copper:

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}

where

\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz

7 0
4 years ago
A metal smith pours 3.00 kg of lead shot at 99oc into 1.00 kg of water at 25oc in an isolated container. what is the final tempe
cluponka [151]
The answer is in attachment.

7 0
4 years ago
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