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Tcecarenko [31]
4 years ago
7

A spring with a spring constant of 50 n/m is compressed 0.2m. if a mass of 0.4kg is connected to the spring, what is the maximum

speed the mass can attain?
Physics
1 answer:
Trava [24]4 years ago
6 0
Use conservation of energy

\frac{1}{2} k x^{2} =  \frac{1}{2} m v^{2}   \\ \\ v =  \sqrt{\frac{k}{m} }  x
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The temperature of a lead fishing weight rises from 26∘c to 38∘c as it absorbs 11.3 j of heat. what is the mass of the fishing w
Nimfa-mama [501]
Given:
ΔT = 38 - 26 = 12°C, temperature change
Q = 11.3 J, heat input
c = 0.128 J/(g-°C), specific heat of lead

Let m =  the mass of the lead.
Then
Q = m*c*ΔT
(m g)*(0.128 J/(g-°C))*(12 °C) = 11.3 J
1.536m = 11.3
m = 7.357 g

Answer:  7.36 g  (2 sig. figs)
8 0
3 years ago
In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002
vichka [17]

Answer:

1.)1.265+or minus 0.0006m

2).0.71%

Explanation:

See attached file

6 0
3 years ago
The earth’s radius is 6.37 × 10 6 m ; it rotates once every 24 hours. What is the earth’s angular speed? Viewed from a point abo
Advocard [28]

Answer:

a) the angular speed of the Earth's rotation is <em>7.272 × 10⁻⁵ rad/s.</em>

<em></em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

<em></em>

c) Earth's speed at a point on the equator is <em>463.23 m/s.</em>

<em></em>

d)  the speed of a point on the earth’s surface halfway between the equator and the pole is <em>231.61 m/s.</em>

Explanation:

a) The angular speed of the Earth's rotation is:

ω = 2π / T

where

T is the period

ω = 2π / (24 hr × (3600 s / 1 hr))

<em>ω = 7.272 × 10⁻⁵ rad/s</em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

c) Earth's speed at a point on the equator is:

v = r × ω

v = (6.37 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 463.23 m/s</em>

d) The radius of the circle in which the point moves is half of Earth's radius. Therefore,

r = 1/2(6.37 × 10⁶ m)

r = 3.19 × 10⁶ m

Therefore, the speed of a point on the earth’s surface halfway between the equator and the pole is:

v = r × ω

v = (3.19 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 231.61 m/s</em>

6 0
4 years ago
Explain a situation in which you can accelerate even though your speed doesn't change
ANTONII [103]
You can accelerate by changing direction, even though you don't speed up/down. Remember that acceleration is a vector so, it has a direction and a magnitude. That's why this works! Hope it helps.
4 0
4 years ago
Read 2 more answers
I'll give brainliest if you give me a good awnser. :)
lbvjy [14]

Answer:

0.5

Explanation:

4 0
3 years ago
Read 2 more answers
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