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Paraphin [41]
2 years ago
8

77. The first law of motion applies to

Physics
1 answer:
Mama L [17]2 years ago
8 0
In the first law, an object will not change its motion unless a force acts on it. So I think the answer is A. Only objects that are moving.
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Please write 4-5 sentences, please!! How would you describe the climate of Colorado? Find 1 piece of evidence from online to def
andreyandreev [35.5K]

Answer:

The climate in colorado is combination of high elevation, midlatitude, and continental interior geography results in a cool, dry, and invigorating climate. The average annual temperature for the state is 43.5 degrees Fahrenheit (F), which is 13.7 degrees below the global mean

Explanation:

4 0
3 years ago
A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A
Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

IMA = \frac{d_i}{d_o}

where

d_i is the input distance

d_o is the output distance

For the pulley system in this problem, d_i = 3.90 m and d_o = 0.975 m, so the IMA is

IMA=\frac{3.90 m}{0.975 m}=4


B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

MA=\frac{F_o}{F_i}

where F_o is the output force and F_i is the input force. For the pulley system in this problem, F_i = 375 N and F_o = 1345 N, so the MA is

MA=\frac{1345 N}{375 N}=3.59


C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

\eta = \frac{MA}{AMA} \cdot 100

Therefore, in this case,

\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%

3 0
3 years ago
Answer?????????????????
oksano4ka [1.4K]

the answer of this question is helium

4 0
2 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a
AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x 10^{24} Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = \frac{GM}{r^{2} }

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g =  \frac{(6.67 * 10^{-11}) (5.972 * 10^{24})  }{5733^{2} }

g = 12.12 m/s^{2}

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}

6 0
3 years ago
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