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3 years ago

9

Who wanna make friends girls onlyyyy

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kirill115 [55]3 years ago

8 0

Lol meeeeeeeeeeee hiii

Liula [17]3 years ago

6 0

Answer:

GO TO H__L..

FILL IN THE BLANKS

<em>FOLLOW</em><em> </em><em>ME</em>

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What are the main activities involved in studying physics?

Firlakuza [10]

The main activity that is involved in studying of physics is the study of natural laws. The study of physics has to do with many aspects of the universe. Physics majorly looks into the natural laws that operate in the universe and describe how they affect matter in relation to time.

7 0

3 years ago

At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's

FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s

Explanation:

Use first equation of motion to find components of velocity at a given time:

$v = u + at$

where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Given:

$u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s$

$v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s$

$v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s$

5 0

2 years ago

Weight on planet Mars

Sunny_sXe [5.5K]

Mars: 0.38

weight = mass x surface gravity

multiplying your weight on Earth by the number above will give you your weight on the surface of Mars

If you weigh 150 pounds (68 kg.) on Earth, you would weigh 57 lbs. (26 kg.) on Mars

5 0

2 years ago

Read 2 more answers

If we assume that, in the African savanna, rainfall and the amount of grass is plentiful, what other factor would BEST explain a

valentina_108 [34]

Diseases, Hunting (sport for humans), predators

8 0

3 years ago

Read 2 more answers

Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o

adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

$F = \sqrt{F_h^2+F_v^2}$

$F = \sqrt{3.36^2+13.98^2}$

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

$\theta =tan^{-1}(\dfrac{F_v}{F_h})$

$\theta =tan^{-1}(\dfrac{13.98}{3.36})$

θ = 76.48°

3 0

2 years ago