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ohaa [14]
2 years ago
13

a 0.500 kg mass is oscillating on a spring with k=330 N/m. the total energy of its oscillation is 3.24 J .what is the amplitude

of the oscillation? (unit=m)
Physics
2 answers:
juin [17]2 years ago
6 0

Answer:

anyone know this or will i have to get my brother

Alekssandra [29.7K]2 years ago
5 0

Answer:

0.140

Explanation:

Give credit to those comment

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A rock hits a window and stops in 0.15 seconds. The net force on the rock is 58N during the collision. What is the magnitude of
nlexa [21]

Answer:

The change in momentum is  \Delta p =   0.7 \ kg\cdot m \cdot s^{-1}

Explanation:

From the question we are told that

    The time taken for the stone to stop is \Delta  t = 0.15 \ seconds

    The net force on the rock is  F =  58 \ N

   

The impulse of the rock can be mathematically represented as

     I  =  F * \Delta t

Substituting values

     I  =  58 * 0.15

    I  =  0.7\  kg * m  * s^{-1}

Now impulse is defined as  the rate at which momentum change

   Hence the change in momentum \Delta p  of the rock is equal to the impulse of the rock

 So  

       \Delta p =  I  =  0.7 \ kg\cdot m \cdot s^{-1}

7 0
3 years ago
what english word does the following- the first two signify a male, the 1st 3 letters support a female, the first four support a
Zarrin [17]

Answer:

lol i know - i was rushing -_-

he= male

her=female

One word with both is <u>heroin</u> but Im not 100% sure

8 0
2 years ago
Are cells made of tisse?
zepelin [54]

Answer:

Yes

Explanation:

Cells make up tissues. Hope this helped

5 0
3 years ago
Read 2 more answers
The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0
2 years ago
A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6
mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87  = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

n=\dfrac{m}{M}

M=Molar mass

n=\dfrac{m}{M}

0.99=\dfrac{28.6}{M}

M=28.88 gm/mol

3 0
3 years ago
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