How did you manage to overcome the difficulties that arose?

400 Kelvin to degrees Celsius

sineoko [7]

Answer:

400-273

127° C

I hope it's helps you

5 0

2 years ago

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a tire has a tread pattern with a crevice every 4.00 cm. each crevice makes a single vibration as the tire moves. what is the fr

koban [17]

the frequency (in hz) of these vibrations if the car moves at 24.2 m/s is 605 HZ .

Calculation :

frequency = $\frac{number of contacts}{second}$

frequency = $\frac{24.2 m/s}{1}*100cm*\frac{1}{4cm}$

= 605 HZ

Frequency describes the number of waves passing through a particular location in a particular time. So if the wave takes 1/2 second to travel, the frequency is 2 per second. If it takes 1/100th of an hour, the frequency is 100 per hour.

Frequency is the number of occurrences of a repeating event per unit time. ... sometimes called time-frequency for clarity,

Learn more about frequency here : brainly.com/question/254161

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5 0

1 year ago

A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a

vovikov84 [41]

This question involves the concept of kinetic energy.

The student's claim is "right".

<h3>Kinetic Energy</h3>

The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

$K.E =\frac{1}{2}mv^2$

where,

K.E = Kinetic energy
m = mass
v = velocity

Therefore,

For the paintball:

$K.E = \frac{1}{2}(4\ g)(90\ m/s)^2$

K.E = 16200 J

For the pellet:

$K.E = \frac{1}{2}(1\ g)(180\ m/s)^2$

K.E = 16200 J

Hence, both paintball and pellet will have same kinetic energy. The student is right.

Learn more about kinetic energy here:

brainly.com/question/12669551

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7 0

2 years ago

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$