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1 year ago

The rate of change of velocity is called:.

Physics

1 answer:

Pavel [41]1 year ago

6 0

Answer: Acceleration

Detailed Explanation:

Acceleration is defined as the rate of change of velocity.

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A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc

saul85 [17]

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

$F = kx$

now we know by Newton's II law

$F = ma$

so here we have

$a = \frac{F}{m}$

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

8 0

2 years ago

Write Newton's 3rd law. How would this law relate to a rocket ship taking off from the earth? Would this law affect the rocket s

irina1246 [14]

Answer:

Part A

Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;

$F_A$ = - $F_B$

Where;

$F_A$ = The action force

$F_B$ = The reaction force

Part B

The law indicates that the force with which a rocket ship uses in taking off from the Earth, $F_A$ is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-) $F_B$

Part C

The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space

Explanation:

6 0

2 years ago

Gggcdfubrdegubtcwftvf y day rx u e HHS’s forget h

aev [14]

Answer:

um . . . yes ?

8 0

2 years ago

Read 2 more answers

To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that

noname [10]

Answer:

a. 2v₀/a b. 2v₀/a

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. = 1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)

= 2v₀/a

4 0

2 years ago

a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame

natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

$\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}$

where;

L = length of the bed

$\mu$ = viscosity

U = superficial velocity

$\epsilon$ = void fraction

dp = equivalent spherical diameter of bed material (m)

$\rho$ = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6 -------------- equation (1)

24A + 576B = 24.1 --------------- equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

B = 0.017014

From equation (1)

12A + 144B = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

$A = \frac{9.6 -144(0.017014}{12}$

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0

2 years ago