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1 month ago

A baseball player runs 27.4 meters from the

Physics

2 answers:

kicyunya [3.2K]1 month ago

8 0

Answer:

Option (3) 3.0 m longer

Explanation:

This problem concerns displacement. Re-examining the scenario:

The first leg to the base = 3 m

The return leg to the base = 3 m

Total distance covered = 6 m

Since the first base is 3 m behind, the net distance = 6 - 3

= 3 m.

kicyunya [3.2K]1 month ago

6 0

6.0 m longer because he ran 3 m out and then ran back 3 m at the end, which makes his net displacement look zero but his actual running distance is 6 m.

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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

ValentinkaMS [3465]

Answer:

The particle's acceleration magnitude and direction is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given:

Mass $m = 1.81\times10^{-3}\ kg$

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Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

To find the particle’s acceleration

The formula for acceleration is

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

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$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, substitute all known values into the formula for acceleration

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

The negative sign indicates the direction is opposite.

Thus, the particle’s acceleration magnitude and direction is $a= 0.3296\ \hat{k}\ m/s^2$

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2 months ago

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Answer:

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