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3 years ago

A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds?

2 answers:

5 0

Initial velocity = 20m/s
Accelerations = 3m/s^2
Time taken = 5s

Using second equation of motion
ut+(1/2)at^2

Where, s is the distance covered by the body
u is initial velocity
a is acceleration and t is time taken by that body

= (20)(5)+(1/2)(3)(5)^2
=100 + 37.5
= 137.5

So, the total distance covered by the body in next 5s would be 137.5m

Arlecino [84]3 years ago

5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

You might be interested in

An elevator has a mass of 1000 Kg. What force is needed to accelerate it upward at a rate of 2 m/s/s?

zubka84 [21]

The force needed to accelerate an elevator upward at a rate of $2 m / s^{2}$ is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as $2 m / s^{2}$ , the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

$\text {Force}=\text {Mass} \times \text {Acceleration}$

$\text { Force }=1000 \times 2=2000 \mathrm{N}=2 \text { kilo Newon }$

So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of $2 m / s^{2}$ .

3 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

3 years ago

Q11) If you were standing at the top of a building and you dropped a rock.

Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0

2 years ago

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

$P_x=rcos\theta$

$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

Velocity at y=150 mm

$150=300sin\theta$

thus $\theta =30^{\circ}$

position of $P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}$

$P_x=259.80 mm$

$P=259.80\hat{i}+150\hat{j}$

Velocity at this instant

$u_x=-rsin\theta =300\times sin30=-150 mm/s$

$u_y=rcos\theta =300\times cos30=259.80 mm/s$

4 0

3 years ago

What is the solution to the problem expressed to the correct number of significant figures?

oksian1 [2.3K]

Answer:

73.72

Explanation:

For this subtraction problem, the answer or solution is expressed to the least precise of the numbers we are trying to subtract.

The least precise number is the number with the lowest significant numbers:

105.4 - 31.681

105.4 has 4 significant numbers

31.681 has 5 significant numbers

So;

105.4

- 31.681

------------------

73.719

----------------

The solution is therefore 73.72

7 0

3 years ago