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Molodets [167]
3 years ago
13

An object with a non-zero speed must be _________.

Physics
2 answers:
pogonyaev3 years ago
8 0
It would be either A or C if its still moving and not stopping
Anika [276]3 years ago
5 0
If you asking about nonzero acceleration this means any type of acceleration greater than zero otherwise Zero acceleration means you are not speeding up or slowing down you are also not changing directions you are merely maintaining a steady speed in one direction.

hope that helps :)
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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
3 0
3 years ago
A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant
Likurg_2 [28]
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3 0
3 years ago
Why do we use atomic models?
vlabodo [156]
Models help us to understand systems and their properties
3 0
2 years ago
You have a 100 ohm resistor. How
sp2606 [1]

Answer:

R2 = 300 Ohms

Explanation:

Let the two resistors be R1 and R2 respectively.

RT is the total equivalent resistance.

Given the following data;

R1 = 100 Ohms

RT = 75 Ohms

To find R2;

Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

RT = \frac {R1*R2}{R1 + R2}

Substituting into the formula, we have;

75 = \frac {100*R2}{100 + R2}

Cross-multiplying, we have;

75 * (100 + R2) = 100R2

7500 + 75R2 = 100R2

7500 = 100R2 - 75R2

7500 = 25R2

R2 = 7500/25

R2 = 300 Ohms

4 0
2 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl
Furkat [3]

Answer:

a) F = 78.606\,N, b) F = 88.911\,N

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m \cdot g \cdot \sin \theta

F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}

F = 78.606\,N

b) The equations of equilibrium are the following:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m\cdot (a + g \cdot \sin \theta)

F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})

F = 88.911\,N

3 0
3 years ago
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