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abruzzese [7]
3 years ago
13

7) HELP ASAP PLZ PLZ HELP ME I don't know if I am right so if I am not right can you plz help me

Physics
2 answers:
artcher [175]3 years ago
6 0
It’s either “B” or “A”
CaHeK987 [17]3 years ago
5 0
I believe that the answer is B
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A 0.l ‑kilogram block is attached to an initially unstretched spring of force constant k = 40 N/m as shown right. The block is d
GalinKa [24]

Answer:

The maximum potential energy of the system is 0.2 J

Explanation:

Hi there!

When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.

The equation of elastic potential energy (EPE) is the following:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretching distance.

The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.

Then:

EPE = 1/2 · 40 N/m · (0.1 m)²

EPE = 0.2 J

The maximum potential energy of the system is 0.2 J

8 0
3 years ago
a carbon atom with 6 proton and 6 neutron in its nucleus is called carbon-12 the carbon atom with 8 neutrons is called what
daser333 [38]

Answer:

carbon-14 must be the answer

4 0
3 years ago
According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-le
Alina [70]
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9

188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
5 0
3 years ago
Một chất điểm DĐĐH có phương trình x=Acos(omega t). Tìm thời gian chất điểm qua vtcb theo chiều dương lần thứ 2?
Arada [10]

Answer:

yyeyrhdhshysyeuhdjxnxmzmakalwooeitjtjdjjs

8 0
3 years ago
What is the passenger's apparent weight at t=1.0s?
Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight =  m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

<u>Apparent Weight = 1035 N</u>

For downward motion:

Apparent Weight =  m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

<u>Apparent Weight = 435 N</u>

4 0
3 years ago
Read 2 more answers
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