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ad-work [718]
3 years ago
7

John attaches a ball to a spring. The diagram below shows what happens. Which option shows the direction of the force of the bal

l on the spring?

Physics
2 answers:
Vika [28.1K]3 years ago
8 0
The answer is downwards because the ball is pulling down the spring due to its weight
Licemer1 [7]3 years ago
6 0

Answer:

c. downwards

Explanation:

took the quiz and got it right

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An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs
OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

E=\dfrac{n^2\pi^2h^2}{8mL^2}

For ground state, n = 1

E_1=\dfrac{\pi^2h^2}{8mL^2}.............(1)

For first excited state, n = 2

40=\dfrac{2^2\pi^2h^2}{8mL^2}.............(2)

Dividing equation (1) and (2), we get :

\dfrac{E_1}{40}=\dfrac{1}{4}

E_1=10\ meV

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0
3 years ago
How much heat is needed to raise the temperature of 9g of water by 17oC?
andre [41]
Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7
6 0
3 years ago
HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
What is the mechanical energy of a 500kg rollercoaster car moving with a speed of 3m/s at the top hill that is 30m high
koban [17]

K.E = 1/2*m*v^2 = 1/2(500)(3)^2 = 2250 J

m*g*h = 500(9.8)(30) = 147000 J

2250 + 147000 = 149250

4 0
3 years ago
What is the acceleration of an object with a constant velocity
valina [46]

Zero.

Acceleration is defined as the change in velocity over time.

Since in your case there is no change, there is no acceleration, so it is zero:

Or in formula: <span>a=<span><span>Δv</span>t</span></span>

Where a=acceleration, <span>Δv</span>=change in velocity and t=time

6 0
3 years ago
Read 2 more answers
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