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2 years ago

A vertically hung spring has a spring constant of 150. newtons per meter. A

1 answer:

7 0

Based on Hooke's law, the force a spring exerts is negative based on the elongation length. Therefore, since a 2.00 kilogram weight exerts 20 Newtons (gravitation acceleration constant = 10), than the spring must pull back with -20 Newtons. This means that -20 = -kx, where k = spring constant and x = elongation length. k = 150, so that means that x = 20/150, or 2/15 meters.

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A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6

Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ = 4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

$v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} = \frac{-461}{92} m/s$

v₃ = -5.011 m/s.

3 0

2 years ago

Read 2 more answers

Sketch a position-time graph for a bear starting

Dmitrij [34]

Explanation:

hopefully that makes sense. the position doesn't change over the 5 seconds, meaning it's stopped but time still continues. then when the slope is negative this shows the bear's position becoming negative (backing up, changing direction).

3 0

3 years ago

Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi

jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

$n_{e}$ = number of electrons = 6.25 x 10⁶

$q_{e}$ = charge on electron = - 1.6 x 10⁻¹⁹ C

$n_{p}$ = number of protons = 7.75 x 10⁶

$q_{p}$ = charge on proton = 1.6 x 10⁻¹⁹ C

Net charge is given as

Q = $n_{e}$ $q_{e}$ + $n_{p}$ $q_{p}$

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0

2 years ago

Which type of friction keeps a pile of rocks from falling apart?

const2013 [10]

The answer is static friction. This is the friction that involves objects that do not move.

6 0

3 years ago

Read 2 more answers

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

3 0

3 years ago