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4 years ago

Plzzz help with the question below!

Physics

1 answer:

KatRina [158]4 years ago

7 0

The answer is place.

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Why is a flower not a good blackbody radiator?

Alecsey [184]

Answer:

A flower emits only visible light

A flower reflects much of the light that hits it

8 0

3 years ago

The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it s

Sholpan [36]

Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

Unstretched length of the spring L is equal to 0.5 feet.

Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds

use the energy conservation equation

$P_A +K_A=P_B+K_B$

Estimate the potential energy , component as position B as below

$P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft$

Estimate the kinetic energy , component as position A as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft$

Estimate the kinetic energy , component as position B as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2$

Substitute 20.5lb- ft for $P_A$

0.5lb-ft for $P_B$

0lb -ft for $K_A$

$0.0777V_2^2 for K_B$

$20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05$

= 16.05ft/sec

7 0

3 years ago

Fill in the blanks ....... combine features of both web apps and native apps

natta225 [31]

Answer:

yvutvuvutvtyvyvtvytvtyvtvytvytvy

3 0

3 years ago

Guys please help me out I’ll give extra points

zvonat [6]

Answer: h = 3.34 m

Explanation:

If the hat is thrown straight up, then at its highest point it has no motion and no kinetic energy. All energy is potential energy

PE = mgh

h = PE/mg = 4.92 / (0.150(9.81)) = 3.34352... ≈3.34 m

8 0

3 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

6 0

3 years ago