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docker41 [41]
3 years ago
12

All matter is made of tiny particles which are in constant...

Physics
2 answers:
jenyasd209 [6]3 years ago
7 0
Which are in constant motion
12345 [234]3 years ago
5 0
Which are in constant motion.
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Need help girliess.
steposvetlana [31]
If I remember correctly, it is the 3rd answer choice.
8 0
3 years ago
Find the Horizontal (x) vector for these forces.
Anestetic [448]

Answer:

See the explanation below.

Explanation:

The force is a vector therefore we can decompose the force into components x & y. as we need the horizontal component of the force, we must use the cosine function of the angle.

F_{1x}=30.8*cos(20)\\F_{1x}=28.94[N]\\F_{2x}=34.3*cos(20)\\\\F_{2x}= 32.23[N]

6 0
2 years ago
What is the numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall?
Nuetrik [128]

Answer:

9.8 m/s/s

Explanation:

The numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall is 9.8 m/s/s. This is called the acceleration due to gravity.

8 0
3 years ago
Read 2 more answers
The initial velocity of a micro van is 15 m/s. It gains a velocity of 40 ms in 10 seconds. Calculate the average velocity and ac
PSYCHO15rus [73]

Answer:

{ \bf{average \: velocity =  \frac{15 + 40}{2}}} \\   = 27.5 \:  {ms}^{ - 1}  \\  { \bf{acceleration =  \frac{v - u}{t} }} \\  =  \frac{40 - 15}{10}  \\  = 2.5 \:  {ms}^{ - 2}  \\  \\ { \tt{second \: qn : }} \\ { \bf{final \: velocity =u + at }} \\ v = 0 + (5 \times 10) \\  = 50 \:  {ms}^{ - 1}

5 0
2 years ago
You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta
Arte-miy333 [17]

Answer:

1)  3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes:    \frac{Distance}{Speed}= \frac{22}{6} =   3.66

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes:    Speed * Time = 34 * 3.66 = 124.44

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : s = u*t + \frac{1}{2}(a * t^2)

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1  = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

4 0
3 years ago
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