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3 years ago

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A man drops a baseball from the top of a building. If the ball is held at a height of 1m before it is dropped, and takes 6.8 sec

onds to hit the ground, how high is the building in meters? (Neglect air resistance)

1 answer:

attashe74 [19]3 years ago

3 0

To solve this problem we will apply the linear motion kinematic equations.

The equation that describes the position as a function of the initial velocity, acceleration and time is given by the relation

$s = v_0 t +\frac{1}{2} at^2$

Here,

$v_0 =$ Initial velocity

t = Time

a = Acceleration, at this case due to gravity

There is not initial velocity then we have that the equation to the given time is

$s = \frac{1}{2} (9.8)(6.8)^2$

$s = 226.8m$

If the ball is held at a height of 1m before it is dropped, we have that the Building height is

$h = 226.8-1$

$h = 225.8m$

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A body of mass 200gram is executing SHM with amplitude of 20mm. The magnitude of maximum force which acts upon it is 0.6N. Calcu

igor_vitrenko [27]

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

$\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2$

Hence,the value of maximum velocity will be 0.3464 m/s².

To learn more about the velocity, refer to the link;

brainly.com/question/862972

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7 0

1 year ago

Read 2 more answers

What was the main view how the world worked geologically prior to the 1960s? It was generally believed that continents and ocean

attashe74 [19]

Answer:

It was generally believed that mountains were produced by vertical forces

Explanation:

The main view of the world worked geologically prior to the 1960s was that the mountains were formed by the vertical forces of nature.

The early people prior to 1960s believed in many different natural phenomenons and they give their own reasons for their occurrence. But later many researchers and geophysicists studied the formation of the earth and came with possible answers to these questions.

Thus the answer is

" It was generally believed that mountains were produced by vertical forces."

3 0

2 years ago

A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

n200080 [17]

Answer:

a

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

b

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

Work done = $m*g*h$

where m is mass

g is acceleration due to gravity

and h is the height

c

Work done in terms of force and distance is = mg

where m is mass of bucket and

g is acceleration due to gravity

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = $\Delta P.E$

W = m*g*h

where g = 9.9 m/ $s^2$

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0

2 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

2 years ago

A beam of light strikes one face of a window pane with an angle of incidence of 30.0°. The index of refraction of the glass is 1

natka813 [3]

Answer:

(a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

Explanation:

Given that,

Angle of incidence = 30.0°

Index of reflection of glass = 1.52

(a). We need to calculate the angle of refraction for the ray inside the glass

Using snell's law

$\dfrac{\sin i}{\sin r}=\mu$

$\sin r=\dfrac{\sin i}{\mu}$

Put the value into the formula

$\sin r=\dfrac{\sin 30}{1.52}$

$r=\sin^{-1}(0.329)$

$r=19.26^{\circ}$

(b). We know that,

The incident ray and emerging ray is equal then the ray will be parallel.

We need to prove that the rays in air on either side of the glass are parallel to each other

Using formula for emerging ray

$\dfrac{\sin e}{\sin r}=\mu$

$\sin e=\sin r\times \mu$

Put the value into the formula

$\sin e=0.3289\times 1.52$

$e=\sin^{-1}(0.499)$

$e=29.9\approx 30^{\circ}$

So, $\sin i=\sin e$

This is proved.

Hence, (a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

6 0

3 years ago