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3 years ago

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1. Определите, используя график, удельную теплоемкость жидкости, если на плавление и нагревание до 65 градусов 1 кг вещества был

о затрачено 200 кДжэнергии. Удельная теплота плавления вещества 95 кДж\кг.

1 answer:

Sholpan [36]3 years ago

6 0

translate it to english

Explanation:

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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

Suppose a scoentist was able to construct a barometer with a liquid being denser than mercury , then how high would the liquid r

ki77a [65]

Answer:

the liquid has less height than the mercury

h_{ liquid} = $\frac{\rho_{Hg} }{\rho_{liqid}} \ h_{Hg}$

Explanation:

The pressure as a function of the height is given by

P = ρ g h

where ρ is the density of the liquid, g the acceleration of gravity and h the height reached by the column of the liquid

In that case they say that the pressure is the standard one that is P = 1.01 10⁵ Pa = 760 mmHg

The first way to give the pressure is in SI units and the second way is the height that the mercury column reaches

In the case of building a barometer with a liquid that has a density greater than that of mercury

ρ_liquid > ρ_Hg

the pressure

P =ρ_lquid g h_liquid

if we have the same pressure

ρ_{Hg} g h_{Hg} = ρ_{liquid} g h_{liquid}

h_{ liquid} = $\frac{\rho_{Hg} }{\rho_{liqid}} \ h_{Hg}$

therefore the liquid has less height than the mercury

7 0

3 years ago

Please explain what is a Energy is

professor190 [17]

Answer:

<h2><u>The capacity or power to do work/ The ability to do work. </u></h2>

Explanation:

such as the capacity to move an object (of a given mass) by the application of force. Energy can exist in a variety of forms, such as electrical, mechanical, chemical, thermal, or nuclear, and can be transformed from one form to another

I hope this help:)

8 0

2 years ago

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

3 years ago

An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi

BARSIC [14]

Answer:

B= 55.6×10^(-7) Tesla

Explanation:

B= μoI/(2πr)

B: magnetic field strength

μo: permeability of free space and is equal to 4π×10^(-7) T.m/A

r: distance from the wire

I : current in the wire

B= (4π×10^(-7)×125)/(2π×4.5)

B= 55.6×10^(-7) Tesla

3 0

3 years ago