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3 years ago

Five men push a stalled car with an average force of 400 N per person. Find the mass of the

Physics

1 answer:

timama [110]3 years ago

4 0

The mass of the car is 2000 kg

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

In this problem, we have:

$a=1 m/s^2$ is the acceleration of the car

Each person applies a force of 400 N, and there are five men, so the total force applied is

$\sum F = 5\cdot 400 N = 2000 N$

Therefore, the mass of the car is:

$m=\frac{\sum F}{a}=\frac{2000 N}{1 m/s^2}=2000 kg$

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

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Is it possible that a speed of 254 and a speed of 100 could be the same speed?

EastWind [94]

Not if both speeds are in the same units.

However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.

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3 years ago

YO CAN ANYONE DO THE BLANK COLUMN AND THE QUESTION PART RQ PLS!!

VikaD [51]

Answer:

stryo: 1

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ice: 1

brick: 2

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Explanation:

d= mass/ total volume

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3 years ago

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

<u>Given:</u>

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

3 0

3 years ago

A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re

arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car = distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a* t^2)/2

distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0

3 years ago

Choose the correct statement regarding the sign conventions for lenses.

Savatey [412]

Answer:

a) false

b) True

c) True

d) False

e) False

Explanation:

a) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

b) True

Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

c) True

d) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

e) False

Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

3 0

3 years ago