1.Anna Traveled 500 m to their house in 600 s. What is the speed of Anna?

Will an object with more mass roll faster down a hill?

inna [77]

Over time, yes. It will over time gain more momentum

5 0

4 years ago

Explaining How Momentum Can Cause

Ghella [55]

Answer:

The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.

Explanation:

6 0

3 years ago

A skier accelerates down the hill at 3m/s2 how fast is he going in 4 seconds

almond37 [142]

Answer:

Explanation: simple kinematics

we suppose that initially vo= 0 so if the skier moves 4s :

vf = vo +at = 0 + 3*4 = 12 m/s

best wishes from colombia

7 0

3 years ago

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a

Naily [24]

Answer:

The angular acceleration of the pencil α = 17 rad·s⁻²

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

τ = I α (1)

W r = I α (2)

The weight is that the pencil has is,

sin 10 = r / (L/2)

r = L/2(sin(10))

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

I = 1/3 M L²

Thus,

mg(L / 2)sin(10) = (1/3 m L²)(α)

α(f) = 3/2(g) / Lsin(10)

α = 3/2(9.8) / 0.150sin(10)

α = 17 rad·s⁻²

Therefore, the angular acceleration of the pencil is 17 rad·s⁻²

3 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