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3 years ago

Why is this a scientific question?

Physics

2 answers:

anastassius [24]3 years ago

8 0

Answer:

Explanation:

Did the quiz

defon3 years ago

7 0

The 3rd one. The question can be tested by a systematic procedure

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Complete the following table with the appropriate units of measurement. Word bank: gram, meter, kelvin, liter, second, grams/cm3

lara31 [8.8K]

mass gram, time sec, temp kelvin, vol liter, dens grams/cm3

5 0

4 years ago

Read 2 more answers

If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p

Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

a = ∆v / ∆t = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force F is proportional to the acceleration a according to

F = m a

where m is the object's mass. Solving for m gives

m = F / a = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0

3 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ $ln[\frac{P_{1} }{P_{2} }]$

= mRT₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵pV = mRT]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = ΔU + W

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

552000 in scientific notation

LUCKY_DIMON [66]

5.52 × 10 to the 5th power (100000) . In scientific notation you need to have a decimal numver times 10 to the power of something so you can divide 552000 by 10 5 times. So in order to get 552000 you need to have 10 to the 4th power and 5.52

4 0

3 years ago

A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th

Fittoniya [83]

Answer:

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k

h) a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

Fe = m a

-k x = ma

a = -k / m x

c) the acceleration is

a = d²x / dt²

If x = A cos wt

v = dx / dt = -A w sin (wt +Ф)

a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

d²x / dt² = -k / m x

We call

w² = k / m

e) substitute to find w

-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

w² = k / m

Angular velocity and frequency are related

w = 2π f

f = 1 / T

We substitute

T = 2π / w

T = 2π √m / k

g) v= - A w sin (wt+Ф)

h) acceleration is

a = - A w² cos (wt+Ф)

8 0

4 years ago