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3 years ago

What does the slope on a speed vs. time graph represent

Physics

1 answer:

steposvetlana [31]3 years ago

5 0

Answer:

It represents the amount of speed per time or (x) for speed (y) for time so the slope would be like y/x

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a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t

Agata [3.3K]

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

5 0

3 years ago

If a star is found directly above the sun on the h-r diagram, what can you conclude about its size?.

hammer [34]

Answer:

It is larger than the Sun.

Explanation:

Brainliest pls

7 0

2 years ago

PLEASE HELP ME SOMEONE

sasho [114]

I think its B or D, most likely D.

7 0

3 years ago

A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change

IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball $v_1=25m/sec$

And velocity after rebound $v_2=-19m/sec$

We have to find the change in momentum

So change in momentum is equal to $=m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec$ ( here negative sign shows only direction )

So option (A) will be correct answer

5 0

3 years ago

The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn

Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

Newton's 2nd law is given as $\Sigma F = ma$ .

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

$\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}$

8 0

2 years ago