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3 years ago

How long would it take for someone to bike to school if the school is 10 km away and the

Physics

1 answer:

damaskus [11]3 years ago

8 0

Answer:

It would take the cyclist 2 hours to cycle to school.

Explanation:

5km/hr for 2 hours would be 10km.

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Katena32 [7]

The net force needed to accelerate a 200 kg piano is 50N

According to Newton's law of motion, the formula for the net force is calculated as:

∑F = ma

Given the following parameters

Mass = 200kg

acceleration = 0.25m/s²

Substitute the given parameters into the formula

∑F = 200 * 0.25

∑F = 50N

Hence the net force needed to accelerate a 200 kg piano is 50N

Learn more here: brainly.com/question/19030143

7 0

3 years ago

Which one of the following crystal structures has the fewest slip directions and therefore the metals with this structure are ge

marta [7]

Answer:

HCP

Extra -> BCC the most, FCC between

Explanation:

4 0

4 years ago

What value is needed to create a motion diagram? a. position c. time interval b. velocity d. both a and c

Fiesta28 [93]

For classical kinematics, you usually graph position versus time, meaning D. This would be a graph where time is on the x-axis and position (can also be treated as displacement if you consider the net distance) is on the y axis.

5 0

4 years ago

Please I need HELP WITH THIS QUESTION!!!

JulsSmile [24]

Oil, grease and dry lubricants

8 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

The frequency changes by a factor of 0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = 0.27

7 0

3 years ago