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Sav [38]
3 years ago
14

Anne and Juan rolled four different objects down a ramp. They measured the distance each object rolled and

Physics
1 answer:
agasfer [191]3 years ago
6 0

Answer:

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A car traveling on the highway at 15 m/s accelerates at 3.0 m/s2 for 5.0 s. What is its final velocity?
Nataly_w [17]

Answer:

The answer to your question is:  vf = 30 m/s

Explanation:

Data

vo = 15 m/s

a = 3.0 m/s²

t = 5 s

vf = ?

Formula

vf = vo + at

Substitution

vf = 15 + (3)(5)

vf = 15 + 15

vf = 30 m/s

3 0
3 years ago
How many oxygen atoms are in the following compound?<br> 8 C120
Novosadov [1.4K]

Answer:

There are 12 oxygen atoms in 8C12O.

6 0
4 years ago
Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second
Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle T_F ,  and the time of the second particle T_S.

Setting the locations equal, we get the following equations to solve for T_F and T_S:

(-1 + T_F) = (-7 + 2T_S)                     Equation 1

(4 - T_F) = (-6 + 2T_S)                        Equation 2

(-1 + 2T_F) = (-1 + T_S)                     Equation 3

Solving these three equations simultaneously we get:

T_F = 2 seconds

T_S = 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of T_F or T_S in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0
3 years ago
To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar
Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

Explanation:

f = \frac{1}{2\pi*\sqrt{C*L}}

a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF  replacing:

f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}

f = 475.14 Hz

b. second smallest use the capacitive 6 uF so:

f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}

f = 1959Hz

c. second largest and largest oscillation first combination so:

Use 4.2 uF

f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{4.2uF*11mH}}

f = 2341.53 Hz

And finally largest oscillation cap in serie so:

C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}

C=2.47 uF

f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}

f =  3053.34 Hz

5 0
3 years ago
You drop a rock off the top of a 100 m tall building. How long does it take to hit the ground?
Vitek1552 [10]
It would mostly depend on its weight
4 0
3 years ago
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