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garri49 [273]
2 years ago
8

An object, moving with a constant velocity, travels 400 m in 25 s. What is its velocity?

Physics
1 answer:
Karolina [17]2 years ago
6 0

v = d/t

= 400/25

= 16 m/s

I'm not sure though

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5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
3 years ago
The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?
Rudik [331]

Answer:

Explanation:

T = 2π√(L/g)

If you increase L to 2L, the period is increased by a factor of √2

T = 3.5√2 ≈ 4.9 s

6 0
3 years ago
HELP
blagie [28]

Answer:

A

Explanation:

hope its right

3 0
3 years ago
7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist
luda_lava [24]

Answer:  5.05 m

Explanation:

4 0
3 years ago
Read 2 more answers
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

Explanation:

We can use the conservation of the angular momentum.

L=mvR

I\omega=mvR

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

(I_{proffesor - stool}+mR^{2})\omega=mvR

Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}      

\omega=0.37 [rad/s]  

I hope it helps you!

5 0
3 years ago
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