Water ______________ easier in the mountains because there is less air pressure.

A student places blocks on a 100cm long see-saw as shown/

Hunter-Best [27]

Answer:

Part 1)

$\tau_1 = 5 \times (0.50) = 2.5 N m$

Part 2)

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Part 3)

$\tau_3 = 1.4 N m$

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

$\tau_1 = 5 \times (0.50) = 2.5 N m$

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Now torque due to weight of the scale is given as

$\tau_3 = 7(0.20)$

$\tau_3 = 1.4 N m$

now torque on left side of scale is given as

$\tau_{left} = \tau_1 + \tau_3$

$\tau_{left} = 2.5 + 1.4 = 3.9 N m$

Torque on right Side is given as

$\tau_{right} = \tau_2 = 4.2 Nm$

Since torque on right side is more so here it will turn and slip over it

8 0

3 years ago

You pull your little sister across a flat snowy field on a sled. Your sister plus the sled have a mass of 22 kg. The rope is at

kati45 [8]

Answer:

Explanation:

mass of sled, m = 22 kg

Force of pull, F = 31 N

distance move, d = 53 m

Angle between force and distance, θ = 40°

The formula for the work done is

W = F x d x Cos θ

W = 31 x 53 x cos 40

W = 1258.6 J

7 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

3 0

3 years ago

If a projectile fired beneath the water, straight up, breaks through the surface at a speed of 13m/s, to what height above the w

OlgaM077 [116]

Well if you had either the velocity or distance traveled i could tell you. But since you haven't all i can say for sure is that the water slowed the bullet down to 13m/s so lets say you knew the distance you would calculate how many meters it traveled and you would have your answer because in this situation, meters (height) =how many seconds spent going into the air.

3 0

3 years ago

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

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