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Aloiza [94]
3 years ago
15

3) Ęplain why muddy water is Heterogeneous mixture

Physics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

muddy water is a heterogeneous mixture, which is Suspension.

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A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

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3 years ago
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The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo
igomit [66]

Suppose car A is moving with a velocity Va, and car b with a velocity Vb,

According the principle of conservation of momentum:

Va x Ma + Vb x Mb = (Ma + Mb) V

V = (Va x Ma + Vb x Mb)/(Ma +Mb)

V = speed of cars after coupling

V = (Va x 20 mg + Vb x  15 mg)/(20 mg + 15 mg)

Put in the values of Va and Vb, and get the V

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3 years ago
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With which of the following statements would Whorf and Sapir agree?
ASHA 777 [7]
Hi!

The answer is <span>B. Language influences how people understand their world.

Hope this helps!

-Payshence xoxo</span>
6 0
2 years ago
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A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
If the driver slammed on the brakes, what could happen to the crate?
stich3 [128]
The crate would slide forward
7 0
3 years ago
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