Ask question

Ask question

All categories

3 years ago

12

What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when release

d?

1 answer:

yawa3891 [41]3 years ago

4 0

To answer this question you have to know weight, shape, and material

You might be interested in

If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will Group of answer c

Sindrei [870]

Answer:

The correct answer is Option A (decrease).

Explanation:

According to Heisenberg's presumption of unpredictability, it's impossible to ascertain a quantum state viewpoint as well as momentum throughout tandem.
Also, unless we have accurate estimations throughout the situation, we will have a decreased consistency throughout the velocity as well as vice versa though too.

Other given choices are not connected to the given query. Thus the above is the right answer.

5 0

3 years ago

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

3 years ago

Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.

amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0

3 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

3 years ago

State two advantages of thermal expansion

Anna71 [15]

Higher Power Efficiency and Eliminates Risk of Compressor Breakdown

6 0

3 years ago