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Sunny_sXe [5.5K]

3 years ago

13

What symbol is used when representing momentum in equations?

1 answer:

Leya [2.2K]3 years ago

6 0

Almost in every Physics books, momentum is denoted by P with small arrow(→) pointing towards right, placed on top of P.
I am guessing you asked about linear momentum.

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Approximately how many atoms are there along a 9.5 cm line

uranmaximum [27]

First convert 5.5 cm to meters.

(5.5 cm / 1) x (m / 100 cm) = 0.055 m

A typical atom is about 1.0E-10 m in diameter; thus:

0.055 m / 1.0E-10 m = 5.5E8 atoms or 550,000,000 end-to-end atoms in 5.5 cm

4 0

2 years ago

Which most likely indicates a chemical change has occurred?

Law Incorporation [45]

green liquid becoming a red liquid

4 0

2 years ago

Read 2 more answers

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

2 years ago

A group of atoms with aligned magnetic poles are known as which of the following?

cestrela7 [59]

A magnetic domain is a group of atoms aligns with magnetic poles. Domains are usually <span>light and dark stripes visible within each grain.</span>

4 0

2 years ago

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A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity

Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

$t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s$

With which we compute the maximum height:

$y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m$

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

$v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s$

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0

2 years ago