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3 years ago

How can you verify the archimedes principle?

Physics

1 answer:

Ipatiy [6.2K]3 years ago

6 0

Answer:

It is found that W1 - W2 loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes' principle.

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Which does not apply to unbalanced forces?

mariarad [96]

Answer:

Net force = 0

Explanation:

Whenever there is an unbalanced force, it shows that the net force doesn’t = 0 because if net force = 0, then everything force is balanced by another force. If there is an unbalanced force, acceleration will occur in the direction the net force is towards.

4 0

3 years ago

A 10kg ball moving at 9.0 m/s strikes a stationary 130kg man on the chest after which the ball bounces back at 5.0 m/s What is t

julsineya [31]

We know that

• The mass of the ball is 10kg.

• The initial velocity of the ball is 9m/s.

• The mass of the man is 130kg.

• The initial velocity of the man is null (stationary).

• The final velocity of the ball is 5m/s.

To find the final velocity of the person, we have to use the law of conservation of momentum.

$p_i=p_f$

Where p = mv. Using all the given information, we have

$\begin{gathered} m_{^{}_{ball}}v_{i_{ball}}+m_{_{man}}v_{i_{man}}=m_{\text{ball}}v_{f_{ball}}+m_{man}v_{f_{man}}_{} \\ 10\operatorname{kg}\cdot9m/s+130\operatorname{kg}\cdot0=10\operatorname{kg}\cdot5m/s_{}_{}+130\operatorname{kg}\cdot v_{f_{man}} \end{gathered}$

Then, we solve for v

$\begin{gathered} 90\operatorname{kg}\cdot m/s=50\operatorname{kg}\cdot m/s+130\operatorname{kg}\cdot v_{f_{man}} \\ \frac{90\operatorname{kg}\cdot m/s-50\operatorname{kg}\cdot m/s}{130\operatorname{kg}}=v_{f_{man}} \\ v_{f_{man}}=\frac{40\operatorname{kg}\cdot m/s}{130\operatorname{kg}}\approx0.31m/s \end{gathered}$ <h2>Therefore, the velocity of the person after the ball bounces off him is 0.31 meters per second, approximately.</h2>

4 0

2 years ago

A parallel-plate capacitor of plate separation d is charged to a potential difference V0. A dielectric slab of thickness d and d

Helen [10]

Answer:

Explanation:

(1.a)

Write the expression for old parallel plate capacitor.

$C = \frac{{{\varepsilon _0}A}}{d}$

The expression of energy stored in the capacitor without any dielectric between the plate is:

$U = \frac{1}{2}C{\left( {\Delta {V_0}} \right)^2}$

Substitute $\frac{{{\varepsilon _0}A}}{d}$ for C

$U = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){\left( {\Delta {V_0}} \right)^2}...(1)$

Supply source is same therefore, the potential difference is constant. So, the energy stored is:

${U_0} = \frac{1}{2}C'{\left( {\Delta {V_0}} \right)^2}$

Write the expression for new parallel plate capacitor.

$C' = \frac{{k{\varepsilon _0}A}}{d}$

Substitute $\frac{{k{\varepsilon _0}A}}{d}$ for C' in expression ${U_0}$

${U_0} = \frac{1}{2}\left( {\frac{{k{\varepsilon _0}A}}{d}} \right){\left( {\Delta {V_0}} \right)^2}...(2)$

From equation (1) and (2).

$\begin{array}{c}\\\frac{U}{{{U_0}}} = \frac{{\left( {\frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){{\left( {\Delta {V_0}} \right)}^2}} \right)}}{{\left( {\frac{1}{2}\left( {\frac{{k{\varepsilon _0}A}}{d}} \right){{\left( {\Delta {V_0}} \right)}^2}} \right)}}\\\\ = k\\\end{array}$

The ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is $\frac{U}{{{U_0}}} = k$

(1.b)

Write the expression for the capacitor $\left( C \right)$ .

$C = \frac{{k{\varepsilon _0}A}}{d}$

The expression for the energy stored by the capacitor is:

${U_0} = \frac{1}{2}C{\left( {\Delta {V_0}} \right)^2}$

Substitute $\frac{{k{\varepsilon _0}A}}{d}$ for C.

${U_0} = \frac{1}{2}\left( {\frac{{k{\varepsilon _0}A}}{d}} \right){\left( {\Delta {V_0}} \right)^2}$

When the distance between the plates is increased and the potential difference between the plates remains constant, then energy stored by the capacitor decreases. Similarly, if the distance between the plates decreases and potential remains constant, then energy stored by the capacitor is increased.

(1.c)

Write the expression for the original charge.

${Q_0} = {C_0}\Delta {V_0}...(3)$

After inserting the dielectric k , the new charge:

$Q = k{C_0}\Delta {V_0}...(4)$

From equation (3) and (4).

$\begin{array}{c}\\\frac{Q}{{{Q_0}}} = \frac{{k{C_0}\Delta {V_0}}}{{{C_0}\Delta {V_0}}}\\\\ = k\\\end{array}$

The charge on the capacitor increases by k

3 0

3 years ago

Một tải có điện trở R = 19ohm đấu vào nguồn điện một chiều có E = 100V,

klio [65]

Answer:

Hindi ko alma yam among twang yan

Explanation:

aorry

6 0

3 years ago

An electron is traveling at 2.0E5 m/s parallel to a uniform electric field of 9.11e-3 N/C strength. The electron is traveling to

anzhelika [568]

Answer:

v = 1.85*10^5 m/s

Explanation:

In order to calculate the speed of the electron after it has traveled 1.8m, you first take into account that the electric field generates a desceleration on the electron, because the direction of the electron and electric field are the same.

You use the Newton second law, to calculate the deceleration of the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19C

m: mass of the electron = 9.1*10^-31kg

E: magnitude of the electric field = 9.11*10^-3N/C

a: deceleration = ?

You solve the equation (1) for a, and replace the values of the other parameters:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(9.11*10^{-3}N/C)}{9.1*10^{-31}kg}\\\\a=1.6*10^9\frac{m}{s^2}$

Next, you use the following formula to calculate the final speed of the electron:

$v^2=v_o^2-2ax$ (2)

v: final speed of the electron = ?

vo: initial speed of the electron = 2.0*10^5 m/s

x: distance traveled by the electron = 1.8m

You solve the equation (2) for v and replace the values of the other parameters:

$v=\sqrt{v_o^2-2ax}=\sqrt{(2.0*10^5m/s)^2-2(1.6*10^9m/s^2)(1.8m)}\\\\v=1.85*10^5\frac{m}{s}$

The speed of the electron after it has traveled 1.8m is 1.85*10^5 m/s

4 0

4 years ago

How can you verify the archimedes principle?​

How can you verify the archimedes principle?