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3 years ago

What is a nonzero net force called

Physics

2 answers:

eduard3 years ago

6 0

Equilibrium.

• When an object is in equilibrium (either at rest or moving with constant velocity), the net force acting on it zero.

olasank [31]3 years ago

5 0

Answer:

Equilibrium. • When an object is in equilibrium (either at rest or moving with constant velocity), the net force acting on it zero.

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Gold-leaf electroscope uses

Kruka [31]

-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)

7 0

4 years ago

Which of the following prefixes represents the largest value

Zigmanuir [339]

C should be right ok

7 0

3 years ago

A dandelion seed floats to the ground in a mild wind with a resultant velocity of 26.0 cm/s. If the horizontal component velocit

stira [4]

Answer:

24 cm/s

Explanation:

Applying

Pythagoras theorem,

a² = b²+c²............. Equation 1

Where a = resultant, b = vertical component, c = horizontal component

From the question,

Given: a = 26 cm/s, c = 10 cm/s

Substitute these values into equation 1

26² = b²+10²

676 = b²+100

b² = 676-100

b² = 576

b = √576

b = 24 cm/s

7 0

3 years ago

Which term describes the image of an object located between a concave mirror and its focal point?

labwork [276]

Answer: I think it is b.

6 0

3 years ago

A probe from earth, with a mass of 106 kg, is in orbit around planet Pud in another solar system. Pud has a radius of 106 m and

DedPeter [7]

Answer:

The acceleration due to gravity in planet Pud is of 9345.7m/s²

Explanation:

Since the probe is describing a circular motion around planet Pud, its centripetal acceleration is given by the formula:

$a_c=\frac{v^{2}}{R}$

Assuming there is no other external forces acting on the probe in the centripetal direction, we have that

$a_c=g_{Pud}=\frac{v^{2}}{R}$

Before we plug in the values, we have to convert the speed from km/s to m/s:

$1\frac{km}{s} =1000\frac{m}{s}$

Finally, using the given values, we can compute g_{Pud}:

$g_{Pud}=\frac{(1000\frac{m}{s}) ^{2} }{107m}= 9345.7\frac{m}{s^{2}}$

In words, the acceleration due to gravity in planet Pud is of 9345.7m/s²

6 0

3 years ago