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3 years ago

6

A heavy box weighing 1000 newtons sits on the floor. You lift upward on the box with a force of 450 newtons, but the box does no

t move. What is the normal force on the box while you are lifting?

2 answers:

Svetllana [295]3 years ago

5 0

The answer for sure is 550

Ex: 1000 is the box
450 is your force
1000-450=550

Anvisha [2.4K]3 years ago

3 0

Answer:

i think its 550

Explanation:

1000-450 is 550

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You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the

bearhunter [10]

Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

$PV=nRT$

For Isothermal process:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

$V_{2}=\frac{V_{1}}{1.1}$ (2)

For Isobaric process:

$\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}}$ (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

$T_{3}=0.8T_{2}$ (4)

For Isochoric process:

$\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}}$ (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

$P_{4}=1.15P_{3}$ (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

$P_{4}=1.15*1.1P_{1}=1.265P1$

This is, the new pressure of the gas increases by 26.5% with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

$V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1$

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

$T_{4}=1.15*0.8T_{1}=0.92T1$

The new temperature decreases 8% with respect to initial volume.

3 0

3 years ago

A concave mirror has a radius of curvature of 1.6m. Find the focal length

Yuki888 [10]

Given,

Radius of curvature of concave mirror = 1.6m

We know that ,

Focal length = radius/2

Hence ,

Focal length of concave mirror = radius of concave mirror /2

=> F = 1.6/2

=> F = 0.8m

Hence the focal length of concave mirror is 0.8 m

5 0

2 years ago

Read 2 more answers

What pushes against gravity in: a main sequence star, a white dwarf, a neutron star, and a black hole? electron degeneracy, neut

Inga [223]

Answer:

heat pressure, electron degeneracy, neutron degeneracy, and nothing

Explanation:

Main Sequence Star: It is a star in which nuclear fusion is happening in the core of the star. Hydrogen molecules fuse together to generate Helium. This nuclear fusion generates outward gas pressure and radiation pressure which balances the inward gravity thus creating an equilibrium which keeps the stars in shape.

White dwarf: It is the end stage of a medium sized star like the Sun. Outer layers of the star are thrown in the form a shell/bubble leaving a small and dense core in the center called as white dwarf. This core consists of carbon and oxygen. Nuclear fusion doesn't occur in the core of white dwarfs. The inward gravity is balanced by the electron degeneracy pressure. Thus these stars will keep on radiating the remaining heat and will turn in to a black dwarf at the end.

Neutron Star: This is the end stage of a supermassive star (1-3 times the mass of the Sun). At the last stage of the life the core collapses. In these stars the inward gravity is so huge that the pressure overcomes the electron degeneracy pressure and crushes together the electron and proton to form neutron. The neutron then stops the collapse and balances the inward gravity.

Black Hole: This is the end stage of a hyper massive stars weighing more than 3 times the mass of the Sun. The inward gravitational force is so huge that even the neutrons are not able to stop the collapse the core. thus the mass of the star collapses into a very small area of immense gravity. There is nothing that can balance this inward gravity.

3 0

2 years ago

Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2

Leya [2.2K]

1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C

3 0

3 years ago

A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig

trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0

3 years ago