On hot sunny days, tire should not be over-inflated,why?

An object of mass 1.5 kg rests on a shelf where it has a gravitational potential energy of 7 joules. An object of mass 4.5 kg is

bija089 [108]

Answer:

C. 21 Joules

Explanation:

We apply the formula to calculate the potential energy (Ep):

Ep=m*g*h

Where:

Ep : potential energy in Joules (J)

m :mass in kilograms (kg)

g acceleration due to gravity (m/s²)

h: height in meters (m)

Calculation of the height (h)

Ep = m*g*h

7 = (1.5 )*(9.8) *(h )

7 = (14.7) (h )

h = 7 / (14.7)

h= 0.476 m

Gravitational potential energy of the second object

Ep = m*g*h

Ep = (4.5 )*(9.8) *(0.476 )

Ep = 21 J

6 0

4 years ago

A construction worker pushes a wheelbarrow with a total mass of 50.0kg. What is the acceleration of the wheelbarrow if the net f

Zarrin [17]

Did you try googling it lol thats what i do if its a problem like that. sometimes there are websites that answer it you just have to look really hard

5 0

3 years ago

1. Describe how the periodic table differentiates between metals and nonmetals.

IgorC [24]

There would be a symbol on names of the elements that would say which state it would usually be in.

there are more metals in the periodic table of elements (currently) if that's what the second question is asking about

7 0

3 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.76 m. After the mass is added, the length is remeasured

ElenaW [278]

Answer:

$1242337 N/m^2$

Explanation:

The cross section area of the cable is

$A = \pi r^2 = \pi * 0.025^2 = 0.002 m^2$

Let g = 9.81m/s2. The stress acting on the cable when mass is added is

$\sigma = F/A = mg/A = 35*9.81/0.002 = 174867 Pa$

The strain when the cable is stretched from 4.76 to 5.43 m is

$\epsilon = \frac{\Delta L}{L} = \frac{5.43 - 4.76}{4.76} = 0.14$

So the young modulus of the cable is

$E = \sigma / \epsilon = 174867 / 0.14 = 1242337 Pa = 1242337 N/m^2$

4 0

3 years ago

Read 2 more answers

White light (400–700 nm) is incident on a 600 line/mm diffraction grating. What is the width of the first-order rainbow on a scr

Tema [17]

Answer:

Δy=0.431m

Explanation:

Diffraction grating with split space d,to find the fringe position ym,we must to find the angle from

dSinα=mλ

A grating with N slits or lines per mm has slit spacing of

d=1/N

d=(1/600mm)

d=1.67×10⁻³mm

For 400nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{400*10^{-9} }{1.67*10^{-6}} )\\ \alpha =13.910^{o}$

And the position of first order lowest wavelength fringe on the screen is:

$y_{1}=Ltan\alpha_{1}\\y_{1}=2tan(13.910)\\ y_{1}=0.49445m$

For 700nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{700*10^{-9} }{1.67*10^{-6}} )\\ \alpha_{2} =24.83^{o}$

And the position of first order highest wavelength fringe on the screen is:

$y_{2}=Ltan\alpha_{2}\\y_{2}=2tan(24.83)\\ y_{2}=0.925595m$

The difference between the first order lowest and highest wavelength fringe is

Δy=(0.925595 - 0.49445)m

Δy=0.431m

6 0

3 years ago