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2 years ago

Terra tosses a 0.20 kg volleyball straight up at 10.0 m/s. how high does it go?

Physics

1 answer:

Sliva [168]2 years ago

5 0

Answer:

5.1 meters

Explanation:

Terra tosses a 0.20kg volleyball up at at a speed of 10 m/s

The height can be calculated as follows

= v^2/2g

= 10^2/2×9.8

= 100/19.6

= 5.1 meters

Hence the height is 5.1 meters

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The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm= $8\times 10^{-2} m$

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid= $8\times 10^{-2} m$

Magnetic field of solenoid

$B=\frac{\mu_0NI}{l}$

Using the formula

$0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}$

Where $\mu_0=4\pi\times 10^{-7}$

$N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns$

6 0

2 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$

Here,

n = Number of electrons

e = Charge of each electron

$n = \frac{q}{e}$

Replacing,

$n = \frac{3.002*10^{-12}}{1.6*10^{-19}}$

$n = 2.44*10^7$

Therefore the number of electrons that reside on the drop is $2.44*10^7$

5 0

3 years ago

How many electrons do the elements of the most stable group have in their outer energy levels

Wittaler [7]

I think it might be 8 electrons

6 0

2 years ago

Read 2 more answers

In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os

GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

$\omega=\dfrac{1}{\sqrt{CL}}$

By putting the values

$\omega=\dfrac{1}{\sqrt{CL}}$

$\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}$

ω = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33 A

I= 6.49 A

I = 6.5 A

E) 6.5 A

3 0

3 years ago

Water flows through a horizontal pipe (from a wider area to narrow area section) and then out into the atmosphere at a speed v1

Darina [25.2K]

Answer:

V = 6.36 m³

Explanation:

For this exercise we will use fluid mechanics relations, starting with the continuity equation.

Let's write the flow equation

Q = v₁ A₁

The area of a circle is

A = π r²

Radius is half the diameter

A = π/4 d²

Q = v₁ π/4 d₁²

Q = π/ 4 15 0.03 2

Q = 0.0106 m3 / s

The volume of water in t = 10 min = 10 60 = 600 s

Q = V / t

V = Q t

V = 0.0106 600

V = 6.36 m³

4 0

2 years ago