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3 years ago

15

A bag of groceries is on the back seat of your car as you stop for a stop light. The bag does not slide. Apply your analysis to

the bag

1 answer:

Romashka [77]3 years ago

7 0

Answer:

This can be the FBD of the bag.

(my bag looks more like a box tho ^^")

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The diagram represents water waves travelling from deep water into an area of much shallower water.

Agata [3.3K]

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;kk

5 0

3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie

elena-14-01-66 [18.8K]

The electric field generated by a point charge is given by:
$E= k_e \frac{Q}{r^2}$
where
$k_e = 8.99 \cdot 10^9 Nm^2 C^{-2}$ is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.

Let's calculate first the electric field generated by the positive charge at that point:
$E_1=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C$
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
$E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C$
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
$E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C$

3 0

3 years ago

Use the worked example above to help you solve this problem. An airboat with mass 4.30 102 kg, including the passenger, has an e

umka21 [38]

Answer:

a) a = 1,865 m / s² and b) t = 8.1 s

Explanation:

a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction

F = m .a

a = F / m

a = 8.02 10² /4.3 10²

a = 1,865 m / s²

b) We use kinematic relationships in one dimension

vf = vo + at

vf = 0 + a t

t = vf / a

t = 15.1 / 1.865

t = 8.1 s

3 0

3 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius $r = 0.330$

m

Force $F = 290$ N

Angular acceleration $\alpha = 0.814 \frac{rad}{s^{2} }$

From the formula of torque,

Γ $= I\alpha$ (1)

Γ $= rF$ (2)

$rF = I \alpha$

Find momentum of inertia $I$ from above equation,

$I = \frac{rF}{\alpha }$

$I = \frac{0.330 \times 290}{0.814}$

$I = 117.56$ $Kg. m^{2}$

Find the momentum inertia of disk,

$I = \frac{1}{2} Mr^{2}$

$M = \frac{2I}{r^{2} }$

$M = \frac{2 \times 117.56}{(0.330)^{2} }$

$M = 2159.045$ Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0

4 years ago