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3 years ago

Rosa records the distance that a toy car rolls and the time it takes to cover the distance. What scientific practice is this?

Physics

2 answers:

suter [353]3 years ago

8 0

collecting data...............................

Sergio039 [100]3 years ago

3 0

Rosa is conducting the scientific practice of observation.

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A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.

marin [14]

Answer: A

Explanation:

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3 years ago

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If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h

nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = $\sqrt{v^2- Vx^2}$ .........................1

put here value

Vy = $\sqrt{35^2- 33^2}$

Vy = 11.66 m/s

and

now we get height

H = $\frac{Vy^2}{2g}$ .............................2

put here value

H = $\frac{11.66^2}{2\times 9.8}$

H = 6.93 m

7 0

3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is 98 N/m.

(B) The period of oscillation is 0.635 s.

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

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3 years ago

Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalp

Contact [7]

The answer is -0.8 m/s. We know acceleration is the average of final minus initial velocity over time (a = (vf-v0)/t). We also know that Force is equal to Mass times Acceleration (F = ma). Using our force equation, we know that the acceleration we get is negative 8.8 (-8.8). The force is acting in the opposite direction of the rugby player, hence the negative sign. From there, plug in that number for a in the velocity equation, and solve for vf, as v0 and t are known. We get 0.8 m/s in the opposite direction that the player was running.

6 0

3 years ago