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3 years ago

PLEASE HELP ME WITH THISSSS

Physics

1 answer:

bija089 [108]3 years ago

7 0

Answer:

she will move in the same direction at the same speed forever.

Explanation:

If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady

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If a participant were holding two different weights in their hands and the jnd for a 10-gram weight was 1 gram, what should the

Nataliya [291]

The jnd for a 100-gram weight, according to Weber's law will be 10 gram.

<h3>What is Weber's law?</h3>

It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.

In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;

= 100 / 10

= 10 gram

Therefore, jnd for a 100-gram weight, according to Weber's law will be 10 grams.

Learn more about weight on:

brainly.com/question/19753744

#SPJ1

7 0

1 year ago

A student is trying to calculate the density of a can. He already knows the mass, but he needs to determine the volume as well.

elena-14-01-66 [18.8K]

Answer:

V=Bh

Explanation:

B h is used for rectangular solids and cylinders

6 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

tatyana61 [14]

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

7 0

3 years ago

A wire connected to red and green terminals on a grey box, which contains a white box with a needle pointing up. The wire passes

harkovskaia [24]

Answer:

A Magnet

Explanation:

Edge 2021

3 0

3 years ago

Read 2 more answers