2 years ago

Calculate the force on an object that has a mass of 12kg and an acceleration of 4m/s2

Physics

1 answer:

Gwar [14]2 years ago

6 0

<span> Force = mass x acceleration. </span>

<span>Force = 12 kg x 4 m/s2 = 48 kgm/s2 = 48 Newtons</span>

You might be interested in

What is the maximum and minimum wavelength of echolocation?

nadezda [96]

I believe the answer is 40,000 Hz to 100,000 Hz

7 0

3 years ago

Read 2 more answers

A conducting sphere of radius r1 = 0.21 m has a total charge of Q = 2.9 μC. A second uncharged conducting sphere of radius r2 =

wolverine [178]

Answer:

what is the anser

Explanation:

4 0

3 years ago

Which of the following is not an essential feature of scientific explanations?

sineoko [7]

Answer:

D I think I might be wrong its been a while scense I did something like that

4 0

3 years ago

Wind erosion can be reduced by _____.

yanalaym [24]

Using land according to its capability. protect the soil surface with some form of cover. control runoff before it develops into an erosive force

4 0

3 years ago

51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago