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melamori03 [73]
3 years ago
11

imagine that you are on a spaceship that is far from the sun or any planet. Explain why a 500 kg piece of equipment would be dif

ficult to move, even though the object would be weightless?
Physics
1 answer:
Mila [183]3 years ago
8 0

Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.

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A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change
Daniel [21]

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

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

Where K indicates spring constant

m indicates mass

For the new time period

T^' = 2\pi \sqrt{\frac{m'}{k} }

Now, we will take 2 ratios of the time period

\frac{T}{T'} = \sqrt{\frac{m}{m'} }

\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }

0.5625 = \sqrt{\frac{0.500}{m'} }

m' = \frac{0.500}{0.5625}

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

7 0
3 years ago
You are driving at 25 m/s with your cruise control on when you see a fallen tree in the road. It takes you 0.30 s to put on the
xxTIMURxx [149]

Explanation:

Given data

velocity v= 25m/s

The time it takes to put on brake t= 0.3s

the distance covered when the  brake was put on is

v=s/t

s= v*t

s= 25*0.3s

s= 7.5m

hence the distance covered is 7.5m

Also the rate of decrease in aceleration is 5m/s^2

we can also calculate the distance covered at this rate

v^2=u^2+2as

25^2= 0+2*5*s

625=10s

divide both sides by 10

s=625/10

s= 62.5m

The total  distance covered between putting on the brakes and decelareation is  7.5+62.5= 70m

Given that the tree is 75m ahead, the car would not hit the tree

3 0
3 years ago
Orange light of wavelength 0.61 µ m in air enters a block of glass with εr = 1.44. What color would it appear to a sensor embedd
77julia77 [94]

Answer:

0.5083\ \mu m

Explanation:

\lambda_0 = Actual wavelength = 0.61\ \mu m

\varepsilon_r = Relative permittivity = 1.44

The observed wavelength in the glass is given by

\lambda=\dfrac{\lambda_0}{\sqrt{\varepsilon_r}}\\\Rightarrow \lambda=\dfrac{0.61}{\sqrt{1.44}}\\\Rightarrow \lambda=0.5083\ \mu m

The wavelength lies in the range of green light.

Hence, the observed color of light is 0.5083\ \mu m

8 0
3 years ago
What is the formula for the speed of a car
Likurg_2 [28]
The formula for speed is s = d/t, where s is speed, d is distance, and t is time. The formula can be applied to all objects, including cars, to find their speed.
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They move in a waves motion

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