Ask question

Ask question

All categories

2 years ago

10

assume that in an unpolished floor a lady us pushing a 40 kg box to the left with an applied force of 192 N. find the sum of for

ce

1 answer:

Vesnalui [34]2 years ago

5 0

The sum of force is 7,680...............

You might be interested in

Which type of stress leads to which type of strain that will lead to which type of surface expression , and cause a reverse faul

kondor19780726 [428]

The answer is C tension; stretching; thinning

7 0

2 years ago

Find the density of an object that has a mass of 5 kg and a volume of 50 cm3

serg [7]

Density = mass/volume. 5000/50= 100 grams per cubic centimeter. No substance on Earth has anywhere near that density. Tops is about 20 for osmium and gold.

3 0

3 years ago

Read 2 more answers

Assume your mass is 84 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb

KATRIN_1 [288]

M, mass=84 kg
height, h=3.9m
gravity, g= 9.8m/s2
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 84•9.8•3.9J
W= 3210.48
Therefore the answer will be 3210.48J.

7 0

1 year ago

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Illusion [34]

<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

v is the velocity in m/s
KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

Based on the problem, the givens are:

KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$

Therefore, the kinetic energy is 9 Joules.

3 0

1 year ago

Sound level B in decibels is defined as

Lelechka [254]

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

3 0

2 years ago