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3 years ago

What would be the force required to accelerate an object that weighs 35 N up at a rate of .60 m/s^2?

Physics

1 answer:

natta225 [31]3 years ago

3 0

Answer: $2.14\ N$

Explanation:

Given

Force required to accelerate the object at the rate of $0.6\ m/s^2$

The weight of the object is $W=35\ N$

Mass of the object can be identified by dividing the weight by acceleration due to gravity

$\Rightarrow m=\dfrac{W}{g}\\\\\Rightarrow m=\dfrac{35}{9.8}\\\\\Rightarrow m=3.57\ kg$

$\therefore \text{Force F=}m\times a\\\Rightarrow F=3.57\times 0.6\\\Rightarrow F=2.14\ N$

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

3 years ago

L'aigua es una mescla de d'oxigen i hidrogen

pshichka [43]

Answer:

Una Mezcla Homogénea es aquella mezcla en la que las sustancias que la forman poseen una combinación uniforme.Son ejemplos de Mezclas Homogéneas: Compuesta

Explanation:

Aire (es una mezcla de gases homogénea formada principalmente por de nitrógeno, oxígeno, vapor de agua, dióxido de carbono...)

Leche (mezcla de agua, carbohidratos, proteínas...)

Bebida alcohólica (mezcla de agua y alcohol etílico)

Acero (mezcla de elementos aleados como el hierro, el carbono y otras sustancias)

Petróleo (mezcla de hidrocarburos)

Agua de mar (mezcla de agua, cloruro sódico y otras sustancias)

Mezcla de agua y sal disuelta

Agua azucarada (mezcla de agua y azúcar)

Aleación metálica (las aleaciones metálicas son mezclas en las que se combinan diferentes metales de una manera homogénea y definida)

Perfume (mezcla de agua y otras sustancias olorosas cuya composición es uniforme)

3 0

3 years ago

A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find

pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

$I=1.41 A$ (current in the wire)

$B=5.61\mu T=5.61\cdot 10^{-6} T$ (strength of magnetic field)

Solving for r, we find the distance from the wire:

$r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m$

4 0

3 years ago

A horizontal spring with spring constant 950 N/m is attached to a wall. An athlete presses against the free end of the spring, c

OLga [1]

Answer:

66.5N

Explanation:

F = kx

Where F = force

K = spring constant

x = compression

Given

K = 950N/m

x = 7.0cm

F = ?

First convert the compression to meters .

7.0cm = 7.0 x 0.01

= 0.07 meters

Therefore

F = 950 x 0.07

= 66.5N

4 0

3 years ago

Hydrogen is found in three different states inside Jupiter, ______, liquid, and liquid ______.

tekilochka [14]

Answer:

gas, metal

Explanation:

The three states of by which hydrogen is found in Jupiter is made up of:

Gaseous hydrogen
liquid hydrogen
liquid metal hydrogen

This is also the same states found in Saturn too.

The pressure inside the largest planet in our solar system is very great.

Hydrogen and helium makes up the entirety of the planet Jupiter.
It has been discovered that inside this planet, hydrogen often occurs as gas, liquid and metal
This is often attributed to the huge amount of pressure in the planet.

7 0

3 years ago