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Nikitich [7]
3 years ago
8

At what step in the scientific method does a scientific propose the problem that he or she wants to solve

Physics
1 answer:
stiv31 [10]3 years ago
8 0

Answer: That would be the first step of the scientific method. After that, the scientist would research the problem and form a hypothesis

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If a car goes down Lake at 30 miles per hour how far will it go in 0.25 hours?
kati45 [8]

Answer:

7.5 miles

Explanation:

0.25 hours=15minutes.

So you do

30 miles=60minutes

x =15minutes

cross multiply

Ans-7.5

7 0
3 years ago
If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
Igoryamba

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

-----------------------

Note: we don't use the mass at all

6 0
3 years ago
Which is the BEST definition of refraction? A) Light or sound waves change direction. B) Light or sound waves bounce off a mediu
melisa1 [442]
The answer to this is A. this is because, refraction with a light or sound wave changing its direction involve propagation,(in which propagation is the change in direction of a light or sound wave)


4 0
3 years ago
Read 2 more answers
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
3 years ago
a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will
sveticcg [70]

Answer:

please give me brainlist and follow

Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

8 0
3 years ago
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