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3 years ago

If you know that the rock fell 4.9 m in 1 s , how far did it drop in the first 0.5 s after you dropped it?

Physics

2 answers:

iren [92.7K]3 years ago

8 0

<u>Answer</u>

1.225 m

<u>Explanation</u>

The formula for getting the distance cover in this case is;

s = ut + 1/2 gt²

Where u⇒intial velocity

t⇒time taken

g ⇒ acceleration due to gravity

But in this case u = 0, t = 1 and s =4.9.

So, the first thin is to calculate the value of g.

S = 1/2 gt²

4.9 = 1/2 g × 1²

g = 4.9 × 2

= 9.8 m/s².

Now calculate the distance after 0.5 s.

S = 1/2 × 9.8 × 0.5²

= 1.225 m

klio [65]3 years ago

7 0

When an object is free-falling, no other force is acting upon it but the gravitational force. Because of this, the equations of motion are simplified. We can determine first the initial velocity:

v = √2gy = √2(9.81)(4.9) = 9.805 m/s

Then, we use this to the equation below:
y = vt + 1/2*at²
y = (9.805)(0.5) + 1/2(9.81)(0.5)²
y = 6.13 m

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A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below

Harlamova29_29 [7]

H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v = the velocity with which the flower pot hits the ground.
Then
v² = 2gh
= 2*(32 ft/s²)*(280 ft)
= 17920 (ft/s)²
v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer: 133.9 ft/s or 91.3 mi/h

5 0

3 years ago

You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j

Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

$E= \frac{1}{2} kx^2$

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

$E=\frac{1}{2} (150) (0.3)^2$

$E=6.75J$

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

#SPJ4

6 0

1 year ago

A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe

lana [24]

Answer: V = 15 m/s

Explanation:

As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately

3 0

3 years ago

In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz

bagirrra123 [75]

Answer:

(a) A = 1 mm

(b) $V_{max}=0.77872 m/s$

(c) $a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A$