Porque al tomar agua y la dejo en mi boca no se baja? Curiosidad xd

A race car has a velocity of 382 km/h to the right. If the car’s mass is 705 kg and the driver’s mass is 65 kg, what force is ne

Lady bird [3.3K]

Answer:

Given: Vi = 382 km/h, Vf = 0 km/h, Mc = 705 kg, Md = 65 kg, Δt = 12

Required: Δx

F = Δp / Δt

= $\frac{(Mc+Md)Vf-(Mc+Md)Vi}{t} \\\\= 6.81 * 10x^{3} N [left]\\\\x=\frac{1}{2} (Vi+Vf)\\ \\ = 637m[right]$

6 0

3 years ago

Is block b speeding up or slowing down once it is set into downward motion and the cat is on block a?

stepan [7]

Answer:

The answer is "Slowing down ".

Explanation:

please find the complete question in the attached file.

In this question, if the block B weight were accounted for by kinetic the friction of frame A, because the blocks pushed at a consistent speed throughout the beginning.

Afterward, on block A, the resistance intensity rises, which allows frames to also be negative, which is defined in the graph, that's why the answer Slowing down is correct.

6 0

3 years ago

QUICK

sashaice [31]

Answer:

The answer is: Increase the Number of Ropes/Pulleys.

Explanation:

A "pulley" is an object that supports a particular movement such as supporting a force's change in direction (provided that the direction is applied to a cord). This is regarded as a "simple machine" consisting of a wheel on an axle. This helps guide cables or ropes.

This machine is used in order to decrease the amount of time it takes to do work, such as lifting of objects.

Remember that "mechanical advantage" refers to the ratio between the force applied to a machine and the force produced by it. So, when it comes to pulleys, the mechanical advantage increases when the number of ropes are also increased. They are directly proportional. This goes the same way with increasing the number of pulleys. Once you increase the number of pulleys, the mechanical advantage also increases.

8 0

3 years ago

A.

Jet001 [13]

Answer:

d one

Explanation:

d one frome my site

7 0

3 years ago

Read 2 more answers

A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m

Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0 (1)

2.070k + Φ = 70.53. (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0

4 years ago