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3 years ago

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En el MCU, la aceleración es: a) constante b) Varía en módulo, direccion, y sentido c) constante solo en el módulo d) constante

solo en dirección y sentido

1 answer:

JulijaS [17]3 years ago

7 0

If you put it into English in the comments I would be more then happy to help you! Thank you!

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Please help! Average speed. Show work!

KiRa [710]

Answer:

3.78 m/s

Explanation:

Recall that the formula for average speed is given by

Speed = Distance ÷ Time taken

Where,

Speed = we are asked to find this

Distance = given as 340m

Time taken = 1.5 min = 1.5 x 60 = 90 seconds

Substituting the values into the equation:

Speed = Distance ÷ Time taken

= 340 meters ÷ 90 seconds

= 3.777777 m/s

= 3.78 m/s (round to nearest hundredth)

3 0

3 years ago

Read 2 more answers

Jack has two boxes: one is 148g and one is 78g. if jack pushes both boxes with the same amount of force which will accelerate fa

kifflom [539]

1) A The 78g
2) C Push on the wagon in the opposite direction as Jack with a force that is the same as Jack is applying.

5 0

3 years ago

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done

sergiy2304 [10]

A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J

3 0

2 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