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2 years ago

8

Two cars approach each other from opposite directions each

1 answer:

olya-2409 [2.1K]2 years ago

5 0

Answer:524 Hz

Explanation:

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After clearing the bar on a high jump, you land softly on a giant mattress. landing on a mattress is more comfortable than landi

madam [21]

Answer:

The right approach is Option b (the force..................exert on you).

Explanation:

Even before you fall on something like a soft object, users eventually slow to a halt. You are still giving away all the downward momentum, but progressively although with small powers, you are doing so.
Although you can get injured by massive powers, this gradual displacement is a positive thing. And that is why you have a mattress you would like to settle on.

The other options given are not connected to the situation described. So, the solution here was the right one.

3 0

2 years ago

What are the characteristics of image when object is between f1 and 2f1 for concave lense?

k0ka [10]

Answer:

The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F

1

and F

1

, its image is formed beyond 2F

2

on the other side of the lens. The image is real, inverted and enlarged.

solution

5 0

3 years ago

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

7 0

3 years ago

There are free body diagrams in the picture, number (#). Which of the diagram(s) shows an object with a net force right?

ryzh [129]

4 only diagram net Force

6 0

3 years ago

Read 2 more answers

An airplane is pushed 22 degrees eastward off its northward course by a jet stream traveling at a speed of 136.73 km/hr. The new

uranmaximum [27]

As the speed of airplane is change due to jet stream

So the net speed is given as

$v_{net} = v_{plane} + v_{stream}$

now we can rearrange it as

$v_{plane} = v_{net} - v_{stream}$

now by the formula of vector difference we have

$v_{plane} = \sqrt{v_{net}^2 + v_{stream}^2 - 2v_{net}v_{stream}cos\theta}$

now plug in all values

$v_{plane} = \sqrt{365^2 + 136.73^2 - 2* 365* 136.73*cos22}[tex]v_{plane} = 243.7 km/hr$

so above is the speed of the plane

3 0

3 years ago