All categories

2 years ago

4. A desk with a mass of 10 kg is pushed across a floor with an acceleration of 4 m/s?. What

Physics

2 answers:

tatiyna2 years ago

6 0

40 N. If you’re using F=MA MxA would be 10x4 which would get you 40 N

Savatey [412]2 years ago

5 0

Answer:

Below

Explanation:

All we need to do for this question is to plug in our values and solve!

Force = mass x acceleration

Force = 10 kg x 4 m/s^2

Force = 40 Newtons or 40 N

If your teacher requires you to round to sig figs it would be :

F = 4 x 10^1 N

Hope this helps!

You might be interested in

Which statement(s) is/are indicative of a covalent bond?

Over [174]

Do you have any answer options?

The most I can help is to tell you that a covalent bond is a type of bond that shares electrons between the bonded atoms.

4 0

3 years ago

An empty parallel plate capacitor is connected between the terminals of a 4.4-V battery and charged up. The capacitor is then di

torisob [31]

Answer:

Explanation:

idk

7 0

3 years ago

Substance A has a higher heat capacity than does substance B, and substance B has a higher heat capacity than does substance C.

svetlana [45]

A < B < C

C will increase faster than B which is faster than A

5 0

3 years ago

The magnitude of vector λa is 5. Find the value of λ, if: a = (−6, 8)

Nata [24]

Given :

$\: \:$

$\color{pink}{\rm \: The \:magnitude \:of \:vector\: λa\: is \: {\bold5\:}}$

$\:$

$\color{green}{\rm \: a = (−6, 8)}$

$\:$

To Find :

$\: \:$

$\orange{ \rm\: value \: of \: \bold{ λ \: }}$

$\: \:$

$\rm \: The \: magnitude \: of \: the \: vector \: a \: is \: ||a|| = \sqrt{(-6)^2+ 8^2)} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm = \sqrt{( 36+64)} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\: \: \:\:= \rm\sqrt{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \rm \underline{\boxed{\red{ \: = 10 \: }}} \green✓$

$\: \:$

$\rm \: Therefore, \: λ = \cancel{\frac{5}{10} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline { \: \boxed{ \red{\: \rm = 0.5\:}}}{ \green ✓ }$