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3 years ago

6

A 54kg robot is moving 9m/s east and collides with a 46 kg robot that is moving 10m/s to the west, what will the final velocity

of the total pile of aluminum and wires be?

1 answer:

Flura [38]3 years ago

7 0

Answer:

0 m/s cause they crashed so they can't move

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What is the direction and speed of an object?

KatRina [158]

Answer:

Explanation:

Velocity by definition means speed and direction of an object. This means it has a value and a positive or minus sign indicating direction. Speed is the absolute value of velocity because there is no direction correlated with speed. If you add a direction, it is then called velocity

7 0

3 years ago

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

Based on the free-body diagram, the net force acting<br> on this firework is

Strike441 [17]

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

4 0

3 years ago

Read 2 more answers

You lift a large bag of flour from the floor to a 2.5m high counter, doing 400J of work in 2 seconds. How much force did you app

oksian1 [2.3K]

<h2>The man have to apply force of 160 N</h2>

Explanation:

The work done to lift the bag of weight mg through height 2.5 m is 400 J

The work done can be found by relation W = mg x h

Thus mg = $\frac{W}{h}$ = $\frac{400}{2.5}$ = 160 N

Therefore the man have to apply the force of 160 N

7 0

3 years ago

Read 2 more answers

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

ololo11 [35]

Answer:

Part a)

$f_B = 290 Hz$

Part B)

percentage increase is

$percentage = 1.38$ %

Explanation:

Part a)

As we know that the beat frequency is

$f_A - f_B = 3$

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

$293 - f_B = 3$

$f_B = 290 Hz$

Part B)

percentage increase in the tension of the string will be given as

$f_A - f_B' = 1$

$f_B' = 292 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

so we have

$T_1 = C (290)^2$

$T_2 = C(292)^2$

so we have

$\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}$

percentage increase is

$percentage = 1.38$

4 0

3 years ago