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2 years ago

What two quantities must stay the same in order for an object to have a constant velocity?

2 answers:

8 0

Velocity is a vector and it has both speed and direction. It takes a force to change direction just as it does to change speed. In order to have a constant velocity the object must maintain a constant direction and speed. Hope this answers the questions. Have a nice day.

Basile [38]2 years ago

8 0

The answer is C: The speed and direction of travel must be constant

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Plz help me to find my science matching...

s2008m [1.1K]

Answer:

fibrous =potato

taproot =radish

stilt =maize and sugar cane

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2 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

2 years ago

An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due

Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

$\displaystyle F_g=G\frac{Mm}{r^2}$

Let's look at Newton's 2nd Law:

$F=ma$

We can set these equations equal to each other:

$\displaystyle G\frac{Mm}{r^2} =ma$

The mass of the second mass (astronaut) cancels out. We are left with:

$\displaystyle G\frac{M}{r^2} =a$

We are solving for the radius of the new planet, so we can rearrange the equation:

$\displaystyle r=\sqrt{\frac{GM}{a} }$

Substitute in our known values given in the problem (G = 6.67 * 10⁻¹¹ ; M = 7.5 * 10²³ ; a = 12).

$\displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }$
$r=2.04 \times 10^6$

The radius of the new planet is ~2.04 * 10⁶ m.

3 0

2 years ago

How is momentum conserved in a Newton's cradle when one steel ball hits the other?

lora16 [44]

Answer:Both balls have momentum to start,and they share it after collision

Explanation:both balls have momentum to start and they share it after collision

7 0

2 years ago

Read 2 more answers

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

$\Delta H_{1}=nC_{p}\times\Delta T$

Put the value into the formula

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

$\Delta H_{1}=nC_{v}\times\Delta T$

Put the value into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Hence, The heat flows into the gas during this two-step process is 120 cal.

7 0

3 years ago