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WINSTONCH [101]
3 years ago
12

A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in

Physics
1 answer:
KatRina [158]3 years ago
6 0

Answer:

37.5 seconds

Explanation:

The given parameters are;

The mass of the block on the spring, m = 2.50 kg

The force with which the loaded spring launches the block, F = 450 N

The change in momentum of the block, Δp = 12.0 kg·m/s

We have;

Let the force with which the block was launched = The net force, F_{NET}

By Newton's second law of motion, we have;

F = F_{NET} = Δp × Δt

Where;

Δt = The time the block is in contact with the spring

Therefore;

\Delta t = \dfrac{F_{NET}}{\Delta p}

By plugging in the values for F_{NET} and Δp, we have;

\Delta t = \dfrac{450 \ N}{12.0 \ kg \cdot m/s} = 37.5 \ s

The time duration the block is in contact with the spring, Δt = 37.5 seconds.

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A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each
jekas [21]

Answer:

884 balloons

Explanation:

Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.

So if pressures reduces from 100 to 1.2, the new volume would be

V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2

The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is

V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3

The number of balloons that 12.5 m3 can fill in is

V_2/V_b = 12.5 / 0.014 = 884

8 0
3 years ago
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista
Nastasia [14]

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

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3 years ago
What happens to gas that is not used
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2 years ago
The 6 strings on a guitar all have about the same length and are stretched with about the same tension.The highest string vibrat
ahrayia [7]

Answer:

B. d(low)=4d(high)

Explanation:

Frequency of a string can be written as;

f = v/2L

Where;

v = sound velocity

L = string length

Frequency can be further expanded to;

f = v/2L = (1/2L)√(T/u) ......1

Where;

m= mass,

u = linear density of string,

T = tension

p = density of string material

A = cross sectional area of string

d = string diameter

u = m/L .......2

m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)

f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))

f = (1/2L)√(T/((p(πd^2)L/4)/L))

f = (1/2L)√(4T/pπd^2)

f = (1/L)(1/d)√(4T/pπ)

Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.

f ~ 1/d

So, if

4f(low) = f(high)

Then,

d(low) = 4d(high)

6 0
3 years ago
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