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Fed [463]
2 years ago
7

Which unit of measurement is best to estimate the volume of a juice box

Physics
1 answer:
stepan [7]2 years ago
8 0

The unit of measurement is best to estimate the volume of a juice box will be kilogram.

<h3>What is volume?</h3>

The capacity of a container like how much quantity of liquid can be filled in that container is called the volume.

The juice bottle is estimated by liter as it contains the liquid. The juice box is a solid cardboard with some volume into which juice bottles are kept. So, the weight will be shown in kilograms.

Thus, the unit of measurement is best to estimate the volume of a juice box will be kilogram.

Learn more about volume.

brainly.com/question/15861918

#SPJ4

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A 4300-N force from a car's engines produces an acceleration of 3.3 m/s2. What is the mass of the car?
sertanlavr [38]

Answer:The net force acting on the car is 3×103 Newtons

Explanation:

5 0
3 years ago
A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increase
ololo11 [35]

As per FBD while its accelerating upwards

we can say that

F_n - mg = ma

here normal force is given as

F_n = 1050 N

W = 840 N

now mass is given as

m(9.8) = 840

m = 85.7 kg

now we will have

1050 - 840 = 85.7 \times a

a = 2.45 m/s^2

Now while accelerating downwards we can say by FBD

mg - F_n = ma

again plug in all values

840 - 588 = 85.7 \times a

a = 2.94 m/s^2

5 0
3 years ago
1. Earth releases about 44-46 Tw of heat, in fact heat can be converted into
Aleksandr [31]

Answer:

  even if it all could be used, it wouldn't be enough

Explanation:

The land area of the US is about 5.45% of the world's area, so the amount of released heat over the area of the US is on the order of 2.4 Tw. Current technology for converting geothermal energy to electricity is about 12% efficient, so the available energy might amount to 0.29 Tw if it could all be captured.

Energy consumption in the US in 2019 was on the order of 0.46 Tw. This suggests that even if <em>all</em> of the thermal energy radiated by the Earth from the US could be turned to useful forms of energy, it would meet only about 60% of the US need for energy.

8 0
3 years ago
A 13561 N car traveling at 51.1 km/h rounds
Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

a_{c} = \frac{v^{2}}{r}

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

F_{c} = ma_{c}

Where:

m: is the mass of the car

The mass is given by:

P = m*g

Where P is the weight of the car = 13561 N

m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg

Now, the centripetal force is:

F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

F_{c} = F_{\mu}

F_{c} = \mu N = \mu P

\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!                

3 0
3 years ago
A 2010 kg space station orbits Earth at an altitude of 5.35×105 m. Find the magnitude of the force with which the space station
Shkiper50 [21]

Answer:

Force, F = 16814.95 N

Explanation:

It is given that,

Mass of space station, m = 2010 kg

Altitude, d=5.35\times 10^5\ m

Mass of earth, m=5.98\times 10^{24} kg

Mean radius of earth, r=6.37\times 10^6\ m

Magnitude of force is given by :

F=G\dfrac{Mm}{R^2}

R = r + d

R=6.37\times 10^6\ m+5.35\times 10^5\ m=6905000\ m

F=6.67\times 10^{-11}\times \dfrac{2010\ kg\times 5.98\times 10^{24} kg}{(6905000\ m)^2}

F = 16814.95 N

So, the force between the space station and the Earth is 16814.95 N. Hence, this is the required solution.

8 0
3 years ago
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