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SIZIF [17.4K]
3 years ago
6

Does a physicist studies cells and life cycles?

Physics
1 answer:
zepelin [54]3 years ago
3 0
Hey there!

No, that would be a biologist. Biologists study biology, or life. They're focused on life cycles, cells, evolution, and so much more.

Physicists focus on physics. They learn about forces, motion, and the study of the universal forces like gravity, momentum, and speed. Just like mathematics, they study patterns and look for new discoveries to be found in the fields of science, physics, and even mathematics.

Hope this helps!
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What type of convergent boundary is the himalayan mountains formed by
Diano4ka-milaya [45]

The Himalayan Mountains formed at a convergence plate boundary between the Eurasian plate and the Indian plate.

8 0
3 years ago
A body of mass 20kg initially at rest is subjected to a force of 40m for 15sec. Calculate the change in k.e
Ede4ka [16]

Explanation:

mass(m)=20kg

velocity(v)=d/t=2.67

k.e=?

now,

k.e=1/2mv^2

=1/2*20*(2.67)^2

=71J

3 0
2 years ago
Read 2 more answers
Which statement is true about Gram negative organisms?
aliina [53]

Answer:

B, C and D are true.

<h3>Explanation:</h3>

A is false because they appear a pale reddish colour not purple.

3 0
3 years ago
A truck is moving with a certain uniform velocity. It is accelerated uniformly by 0.75 m/s^2. After 20 seconds , the velocity be
scoundrel [369]

Answer:

Vi = 5 m/s

Explanation:

let (a) acceleration = 0.75 m/s²

(t) time = 20 seconds

Vf = final velocity = 72 km/hr  (convert to m/s to units consistency = 20 m/s)

find Initial velocity (Vi)

       Vf - Vi

a =  -----------

             t

Vi = Vf - (a * t)  = 20 - (0.75 * 20)

Vi = 5 m/s

5 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
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