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2 years ago

5

A man is holding a 5kg weight ball. What is the weight of the ball?

1 answer:

jonny [76]2 years ago

3 0

Answer:

1000kg

Explanation:

I'm smart I am a genius I know everything so I know it is 1000kg

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I need help please only way to put my grade up !!!!!! Would appreciate it !!! Someone who’s good at this

motikmotik

Answer:

Total energy = 1000J

KE = 500J

PE = 500J

Explanation:

As you may know, the equation for gravitational potential energy is mgh (weight x height)

If the skateboard is halfway down, that means it is at half the height. As the skateboard speeds up (as it goes downward), the potential energy becomes kinetic energy. Since it has 500J of kinetic energy at half way down, it means it had double that amount of Potential energy at the top (1000J). Since half of that became kinetic energy, there is only 500J of PE left.

Total energy = KE + PE = 1000J

KE = 500J

PE = 500J

3 0

3 years ago

Ninety percent of stars in the solar system are :

sergiy2304 [10]

90% of stars in the solar system are main sequence stars.

3 0

3 years ago

Identify the following items as scalar or vector.

Salsk061 [2.6K]

Answer:

Speed - scalar

Velocity - vector

Displacement - vector

Distance - scalar

Measurement - scalar

Measurement and direction - vector

60 m north - vector

100 m west - vector

200 m/s - scalar.

7 0

4 years ago

In a double-slit experiment, the third-order maximum for light of wavelength 490 nm is located 15 mm from the central bright spo

jeka94

Answer:

The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

Explanation:

Given that,

Wavelength = 490 nm

Distance y= 15 mm

Length L=1.6 m

New wavelength = 670

We need to calculate the distance from the central bright spot

Using formula of distance

$y= \dfrac{m\lambda L}{d}$

$d=\dfrac{m\lambda L}{y}$

Put the value into the formula

$d=\dfrac{3\times490\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=156.8\times10^{-6}\ m$

$d=156.8\times10^{-3}\ mm$

We need to calculate the distance from the central bright spot for new wavelength

Using formula of distance

$d=\dfrac{m\lambda L}{y}$

Put the value into the formula

$d=\dfrac{2\times670\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=142.9\times10^{-6} m$

$d=142.9\times10^{-3}\ mm$

Hence, The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

8 0

3 years ago

Just helping as usual what about you

7nadin3 [17]

Answer:

same i just love helping people out

god bless you have a blessed day

Explanation:

6 0

3 years ago

Read 2 more answers