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3 years ago

HELP ME PLSS!! Will mark brainiest

Physics

2 answers:

VikaD [51]3 years ago

5 0

1.skull

2. bone marrow

3. homeostasis

nataly862011 [7]3 years ago

3 0

Answer:

1.Skull

2.Bone marrow

3.Homeostasis

Explanation:

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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

3 years ago

2CO + 2NO → 2CO2 + N2 is it balannced

natita [175]

If the same atoms appear on both sides, then it's balanced.

In this reaction, there are 4 Oxygens, 2 Carbons, and 2 Nitrogens on each side. So numerically, <em>it's balanced</em>. But I don't know enough chemistry to say whether the reaction is possible.

4 0

3 years ago

Read 2 more answers

In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the

Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0

3 years ago

I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT

ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR = E- V

IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

$R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms$

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0

3 years ago

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

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

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

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

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

3 years ago