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3 years ago

Distinguish between speed and velocity.

2 answers:

8 0

Sometimes they are the same in physics. it depends how the question is worded but speed comes before velocity. If you take the derivative of speed you will end up with its velocity. The same goes when you take the integral of velocity.

Alona [7]3 years ago

6 0

The most simple difference between velocity and speed is: velocity has both magnitude and direction, while speed only contains magnitude. Hope that helps you.

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A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67

aliya0001 [1]

Acceleration =velocity /time
=42/7
=6

5 0

4 years ago

Read 2 more answers

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

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

6 0

3 years ago

What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

$V_{e}=\sqrt{\frac{2GM}{R}}$

Where:

$V_{e}$ is the escape velocity

$G=6.67(10)^{-11} Nm^{2}/kg^{2}$ is the Universal Gravitational constant

$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

$V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}$

Finally:

$V_{e}=3633.86 m/s \approx 3.63 km/s$

7 0

3 years ago

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l

Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

Wₐ = 1500 lb

W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

W = mg

m = W / g

mₐ = Wₐ / g

m_b = W_b / g

mₐ = 1500/32 = 46.875 slug

m_b = 125/32 = 35,156 slug

Let's reduce to the english system

v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

p_f = (mₐ + m_b) v

the moment is preserved

p₀ = p_f

mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

v = $\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}$

we substitute the values

v = $\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6$

v = 0.559 v₀ₐ - 26.40 (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

M = mₐ + m_b

M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

K₀ = ½ M v²

final point. When they stop

K_f = 0

The work is

W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

N-W = 0

N = W

the friction force has the expression

fr = μ N

we substitute

-μ W x = Kf - Ko

-μ W x = 0 - ½ (W / g) v²

v² = 2 μ g x

v = $\sqrt{ 2 \ 0.750 \ 32 \ 17.5}$ Ra (2 0.750 32 17.5

v = 28.98 ft / s

3 0

2 years ago