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2 years ago

What distinguishes a nebula and a star?

1 answer:

4 0

What distinguishes a nebula and a star is that a nebula is a cloud of gas and dust in outer space, visible in the night sky either as an indistinct bright patch or as a dark silhouette against other luminous matter. And a star is a type of astronomical object consisting of a luminous spheroid of plasma held together by its own gravity.

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Luis wanted to know which of his 4 toy cars is the fastest. He conducted an experime and recorded his results in the table shown

Murrr4er [49]

The dependent variable was the time and the independent variable was thecars

6 0

3 years ago

if there are infinite universes is it possible that there is a universe where 1+1=3 or 4 or something

zhenek [66]

Answer:

noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo i said nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

Explanation:

8 0

2 years ago

a ball rolls along the floor with a constant velocity of 3 m/s. How far will it have gone after 234 seconds?

sergij07 [2.7K]

Answer:

The ball travel for 702m

Explanation:

distance = speed × time

speed = 3m/s

time = 234s

distance = 3 × 234

= 702m

7 0

2 years ago

Read 2 more answers

Vascular plants can be divided into three categories. These categories are

Evgesh-ka [11]

I believe the answer here is C).seedless, seed (nonflowering), and seed (flowering).

6 0

2 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

2 years ago