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Roman55 [17]
3 years ago
12

Ajar can hold 60 mL of liquid. It is 10 cm high and 2 cm wide. The length of the jar is:

Physics
2 answers:
ZanzabumX [31]3 years ago
8 0
60:2=30 despues 30:10= 3
marin [14]3 years ago
5 0

Answer:

3cm

Explanation:

From the question, we obtained the following:

Volume = 60mL = 60cm^3

Height = 10cm

Width = 2m

Length =?

Volume = length x width x height

Length = volume / width x height

Length = 60/(2 x 10) = 60/ 20

Length = 3cm

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A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall
kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

F = q(\vec v \times \vec B)

so here we have

F = qvB sin90

since force is perpendicular to the velocity so here it must be centripetal force

here we have

\frac{mv^2}{R} = qvB

so we have

R = \frac{mv}{qB}

R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}

R = 4.06 cm

so here it will move in circle with radius 4.06 cm

8 0
3 years ago
Use the Venn Diagram to compare and contrast solar and lunar eclipses.
Sergio [31]
From our perspective on Earth, two types of eclipses <span>occur: </span>lunar<span>, the blocking of the </span>Moon<span> by Earth's shadow, and </span>solar, the obstruction of the Sun by the Moon<span>. ... When Earth passes directly </span>between<span> Sun and </span>Moon<span>, its shadow creates a </span>lunar eclipse<span>.</span>
5 0
3 years ago
The graph shows a car's velocity over time. During which time period does the car have the greatest acceleration?
Oksana_A [137]

Answer:

Between 0 and 1 seconds (B)

Explanation:

The velocity of the car over time is represented by the line graphed here

the steeper the line, the greater change in velocity that occurred in a given time frame.

The steepest portion of the line is between 0-1 seconds, which means that the greatest rate of change occurred between 0-1 seconds.

(acceleration is the rate of change)

5 0
2 years ago
Read 2 more answers
A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
3 years ago
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