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Musya8 [376]
2 years ago
13

1. A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?

Physics
1 answer:
faltersainse [42]2 years ago
6 0

Answer: By 47

Explanation: subtract 310 and 263

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5. How does privatization contribute to productivity?
tangare [24]

Answer:

increased competition

Explanation:

3 0
2 years ago
6. An object accelerates from rest to 70 m/s in 3.5 s. What is the acceleration of the object?
sweet-ann [11.9K]

Answer:

The acceleration of the object is 20 meters per second square = 20 m/s^2

Explanation:

Recall that acceleration is defined as the change in velocity divided the time it takes for the change. Therefore , if the object accelerates from rest (zero velocity) to 70 m/s , the change in velocity is (70 m/s - 0 m/s = 70 m/s)

which divided by the 3.5 seconds it took for the change, gives:

acceleration = (70 m/s  /  3.5 s ) = 20  m/s^2

4 0
3 years ago
At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himsel
Illusion [34]

Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

Fₙ = 20 – 15

Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

7 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what
pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

#SPJ4

8 0
2 years ago
A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has
lara [203]

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = \sqrt{(\frac{v_x}{\omega })^2+y^2 }  = \sqrt{(\frac{35}{2\times \pi  \times y})^2+6.2^2 }  =  6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

5 0
3 years ago
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