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3 years ago

Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?

Physics

1 answer:

klasskru [66]3 years ago

6 0

Explanation:

It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.

We need to find the speed before and after the ball bounce.

Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

$\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s$

Let v is the final speed when it bounces to a height of 1.5 m. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s$

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.

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Water is moving at a velocity of 2.00 m/s through a hose with an Internal diameter of 1.60 cm. What is the volume flow rate at t

Leya [2.2K]

Answer:

15 m/s

Explanation:

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3 years ago

A platinum ball weighing 100 g is removed from a furnace and dropped into 400 g of water at 0 degree C. If the equilibrium tempe

kakasveta [241]

Answer:

T = 1010 degree Celsius

Explanation:

mass of ball (Mb) = 100 g

mass of water (Mw) = 400 g

temp of water = 0 degree

specific heat of platinum (C) = 0.04 cal/g degree celsius

we can calculate the temperature of the furnace from the equation before

Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)

100 x 0.04 x ( T - 10) = 400 x (10 - 0)

4 (T - 10) = 4000

T - 10 = 1000

T = 1010 degree Celsius

3 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

elena-14-01-66 [18.8K]

The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

c is speed of light
E₀ is the peak electric field
B₀ is the peak magnetic field

B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

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2 years ago

Given: F = k· m. g<br> Solve for "k"

gulaghasi [49]

Answer:

$F = kmg \\ k = \frac{F}{mg}$

8 0

3 years ago

Read 2 more answers

M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.

MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is $6.49\times10^6m/s$

<h3>How to calculate the speed of the electron?</h3>

We know, that the energy of the system is always conserved.

Using the Law of Conservation of energy,

$\triangle K+\triangle$ U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

$(\frac{1}{2} mv^2)+(-q\triangle V)$ =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let $v_f$ and $v_i$ represent the final and initial speed.

Here, $v_i$ =0

Solving for $v_f$ , we get:

$v_f=\sqrt{\frac{-2q\triangle V}{m} }$

$=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }$

=6.49 $\times10^6$ m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

4 0

1 year ago