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loris [4]
3 years ago
5

A plant dying after being exposed to poison represents a physical change. True False

Physics
2 answers:
nlexa [21]3 years ago
8 0

Answer:

chemical change as death is irreversible

nasty-shy [4]3 years ago
5 0

Answer:

It does not represent a physical change

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A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of th
Contact [7]

Answer:

C. 450v

Explanation:

Using

Voltage= B*distance of separation*velocity

3mm x 0.3T x 5E5m/s

= 450v

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3 years ago
2. In these diagrams, the sunlight is coming from the left, as shown by the arrows. Which
RideAnS [48]

Answer:

Explanation:

a) A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.

3 0
3 years ago
A 0.70-kg disk with a rotational inertia given by MR 2/2 is free to rotate on a fixed horizontal axis suspended from the ceiling
Setler [38]

Answer:

The force will be "9.8 N".

Explanation:

The given values are:

mass,

m = 0.7 kg

M = 2

g = 9.8

Now,

⇒  \tau = T \alpha

then,

⇒  \frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R

⇒  \frac{1}{2}m \ a_t=m(g-a_t)

⇒  a_t=\frac{2g}{(\frac{m}{M} +2)}

On substituting the values, we get

⇒      =\frac{2\times 9.8}{\frac{0.7}{2} +2}

⇒      =8.34 \ m/s

hence,

⇒  T=mg+M(g-a_t)

On substituting the values, we get

⇒      =0.7\times 9.8+2(9.8-8.34)

⇒      =6.86+2(1.46)

⇒      =6.86+2.92

⇒      =9.8 \ N

5 0
3 years ago
Calculate the speed of sound in a string that has a tension of 100 N and a linear mass density of 0.0001 kg/m
Svetradugi [14.3K]
You will need to divide it.
8 0
2 years ago
Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If
frez [133]

Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2  

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2  

v2 = v1×A1/A2  

and  

r2 = 2×r1

then:

A2 = 4×A1  

so,

v2 = (v1)/4  

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2  

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)  

     = 9.75×10^4 Pa

    = 9.8×10^4 Pa

Therefore, the pressure in the wider section is 9.8×10^4 Pa

5 0
3 years ago
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