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3 years ago

14

What tool could help a biologist study the movements of cells? digital camera with a zoom lens a satellite with a digital camera

a television with a remote control a microscope with a video camera

2 answers:

Annette [7]3 years ago

4 0

I think your answer would be (D) microscope with a video camera
Hope i helped!

Mandarinka [93]3 years ago

4 0

Answer:

a microscope with a video camera

Explanation:

gg Dva style

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A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper pla

Hatshy [7]

Answer:

The time rate of change of the electric field between the plates is $\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$

Explanation:

From the question we are told that

The radius is $r = 2.8 \ cm = 0.028 \ m$

The distance of separation is $d = 1.1 \ mm = 0.0011 \ m$

The current is $I = 5 \ A$

Generally the electric field generated is mathematically represented as

$E = \frac{q }{ \pi * r^2 \epsilon_o }$

Where $\epsilon_o$ is the permitivity of free space with a value

$\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

So the time rate of change of the electric field between the plates is mathematically represented as

$\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }$

But $\frac{q}{t } = I$

So

$\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }$

substituting values

$\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }$

$\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Varvara68 [4.7K]

The answer to your question is,

4 kilometers north

-Mabel <3

6 0

2 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

stira [4]

Answer:

See explanation

Explanation:

We have a mass $m$ revolving around an axis with an angular speed $\omega$ , the distance from the axis is $r$ . We are given:

$\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]$

and also the formula which states that the kinetic rotational energy of a body is:

$K =\frac{1}{2}I\omega^2$ .

Now we use the kinetic energy formula

$K =\frac{1}{2}mv^2$

where $v$ is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

$v=\omega r$

After replacing in the previous equation we get:

$K =\frac{1}{2}m(\omega r)^2$

now we have the following:

$K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2$

therefore:

$mr^2=I$

then the moment of inertia will be:

$I = 13*(0.5)^2=3.25 [Kg*m^2]$

3 0

3 years ago

If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.

kodGreya [7K]

Answer:

$\Delta H=687.4 J$

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

$\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J$

$\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J$

Therefore, the required heat is:

$\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J$

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0

2 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago