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zaharov [31]
3 years ago
10

Both cars start from the same point. Which describes the motion shown?

Physics
2 answers:
Alekssandra [29.7K]3 years ago
8 0
C) The longer the line, the greater the magnitude of the vector. As for the direction, just think of a compass.




timama [110]3 years ago
7 0

C: one car travels south at 4 m/s. The other car travels east at 2 m/s

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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?
poizon [28]

Answer:

\boxed {\boxed {\sf 7.84 \ Joules}}

Explanation:

The formula for potential energy is:

PE=m*g*h

where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m

Substitute the values into the formula.

PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)

Multiply the first two numbers.

  • 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
  • If we convert the units now, the problem will be much easier later on.
  • 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

PE=(3.92 \ N )(2 \ m)

Multiply.

  • 3.92 N* 2 m=7.84 N*m
  • 1 Newton meter is equal to 1 Joule (this is why we converted the units).
  • Our answer is equal to<u> 7.84 Joules.</u>

PE=7.84  \ J

6 0
2 years ago
Read 2 more answers
How many cubic feet are in a 55-gallon drum?
san4es73 [151]
1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches

                       (55 gal) x (231 in³/gal) x (1 ft³/1728 in³)

                   =    (55 x 231 / 1728)  ft³

                   =         7.352 cubic feet       (rounded)  
5 0
3 years ago
If a baseball has a zero velocity at some instant, is the acceleration of the baseball necessarily zero at that time? Explain -
ipn [44]

Answer:

No, not necessarily

Explanation:

If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.

An example of an object that has zero velocity but non-zero acceleration:

If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 m/s^{2} rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 m/s^{2}.

3 0
3 years ago
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0
2 years ago
A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v
kifflom [539]

Here we will say that there is no external torque on the system so we will have

L_i = L_f

here we know that

L_i = I_1\omega_1

where we know that

I_1 = \frac{2}{5}mr^2

Also we know that

I_2 = \frac{1}{12}mL^2

initial angular speed will be

\omega_1 = 2\pi(2rev/s) = 4\pi rad/s

now from above equation

\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega

0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega

0.452 = 0.1875 \omega

now we have

\omega = 2.41 rad/s

so final speed will be 2.41 rad/s

6 0
2 years ago
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