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3 years ago

Why will a change in velocity have a greater affect on KE than a change in mass?

Physics

2 answers:

Sever21 [200]3 years ago

5 0

Answer: the answer is corect of what everone else said

Explanation:

lys-0071 [83]3 years ago

5 0

Answer:

..: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

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A copper sphere 10 mm in diameter is dropped into a 1-m-deep drum of asphalt. The asphalt has a density of 1150 kg/m3 and a visc

kvasek [131]

Answer:

t = 1964636.542 sec

Explanation:

Given data:

sphere diameter is 10 mm

Density is 1150 kg/m^3

viscosity 105 N s/m^2

We knwo that time taken by sphere can be calculated by following procedure

$\tau = \mu \frac{du}{dy}$

$\frac{F}{A} = \mu \frac{du}{r}$

$\frac{\rho_C -\rho_{asphalt} gv}{2 \pi rL} = 10^5 \frac{du}{r}$

Solving for du

$du = \frac{ (8933 - 1150) 9.81 \frac{4}{3} \pi (10\times 10^{-3})^3}{2\pi \times 1\times 10^5}$

$du = u = 5.09\times 10^{-7}$

$u = \frac{1}{t}$

$t = \frac{1}{5.09\times 10^{-7}} = 1964636.542 sec$

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4 years ago

what is the mass of a in the cylinder if the cross sectional area has 0.4cm square and the vertical length of the liquid is 2cm

Leni [432]

Explanation is in a file

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3 years ago

How do you calculate percentage uncertainties???

ch4aika [34]

I is just the uncertainty over the actual number multiplied by 100.

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36.1( +/- 0.1)

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