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DochEvi [55]
3 years ago
10

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced
Physics
1 answer:
charle [14.2K]3 years ago
4 0

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

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What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes
mr Goodwill [35]
The current is defined as the amount of charge Q that passes through a given point of a wire in a time \Delta t:
I= \frac{Q}{\Delta t}
Since I=500 A and the time interval is
\Delta t=4.0 min=240 s
the charge is
Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C

One electron has a charge of q=1.6 \cdot 10^{-19}C, therefore the number of electrons that pass a point in the wire during 4 minutes is
N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23} electrons
3 0
3 years ago
In which mechanical test is a specimen deformed with a gradually increasing load that is applied uniaxially along the long axis
IRINA_888 [86]

Answer:

Compression Test

Explanation:

The Specimen is undergoing a compression test. It is similar to tensile test with the difference that the force is compressive and applied along the direction of stress. Both Tensile and compression tests are performed on Universal Testing machine. Compression test is done to determine the product's reaction when it is compressed, squashed and crushed.

7 0
3 years ago
A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical
loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

PE =0.92414J

KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

3 0
3 years ago
Where are the nodules of a legume found?
Sveta_85 [38]

Answer:

The answer is roots.

Explanation:

Root nodules are found on the <em>roots of plants</em>, primarily legumes, that form a symbiosis with nitrogen-fixing bacteria. so it is roots

6 0
3 years ago
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