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3 years ago

All of these are very important parts of studying physics EXCEPT

Physics

2 answers:

Sergeeva-Olga [200]3 years ago

5 0

Answer:

4. deducing and applying natural laws

murzikaleks [220]3 years ago

4 0

4 since i am right.

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If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B

postnew [5]

Let B = a i + b j + c k. Then the cross product of A = i + 2j - k with B is

A × B = ( i + 2j - k ) × ( a i + b j + c k )

A × B = a ( i × i ) + 2a ( j × i ) - a ( k × i )

… … … + b ( i × j ) + 2b ( j × j ) - b ( k × j )

… … … + c ( i × k ) + 2c ( j × k ) - c ( k × k )

A × B = 0 - 2a k - a j

… … … + b k + 0 + b i

… … … - c j + 2c i - 0

A × B = (b + 2c) i + (-a - c) j + (b - 2a) k

So we have

3 i - j + 5 k = (b + 2c) i + (c - a) j + (b - 2a) k

which gives us the system of equations,

{ b + 2c = 3

{ -a - c = -1

{ -2a + b = 5

Solve for a, b, and c.

• Eliminate c from the first two equations:

(b + 2c) + 2 (-a - c) = 3 + 2 (-1)

-2a + b = 1

But -2a + b = 5, and 5 ≠ 1, so there is no such vector B that satisfies the cross product!

8 0

3 years ago

At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she tra

Ad libitum [116K]

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

$x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m$

7 0

3 years ago

An object initially at rest experiences an acceleration of 9.8 m/s2. How much time will it take to achieve a velocity of 58 m/s?

frez [133]

5.91(approx) seconds just divide velocity by acceleration

4 0

4 years ago

A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in

Paraphin [41]

Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

$v_{av}=\frac{d}{t}$

where,

$v_{av}$ = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

$v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr$

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

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3 years ago

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In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak

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How can we experimentally verify newton's laws?

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4 years ago

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