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liq [111]
3 years ago
9

What is the name of Newton's first law of motion? A. law of inertia B. law of acceleration C. law of objects at rest D. law of o

pposing forces
Physics
2 answers:
Ghella [55]3 years ago
8 0

the answer would be law of inertia

givi [52]3 years ago
3 0
The answer is a.......
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Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche
DaniilM [7]

Answer:

R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}

Explanation:

Using kinematics equations:

\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t

Use \Delta y = 0 due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

t=\frac{2v_{0y}}{g}

Use the expression in the first equation to have

R = \frac{2v^2 \cos\theta\sin\theta}{g}

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

5 0
3 years ago
Help please due by 3:00
FrozenT [24]

Answer:

  zero

Explanation:

Acceleration is a measure of the rate of change of velocity. If the velocity is unchanging, its rate of change is zero.

The acceleration is zero.

8 0
3 years ago
Dark areas on the surface of the Sun that exhibit intense magnetic activity are
svp [43]
I think it's A) Sunspots. I hope this helps:)    
6 0
3 years ago
Read 2 more answers
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
DETERMINE THE BCD EQUIVALENT OF 1100111110101001
Solnce55 [7]

Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

6 0
2 years ago
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