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3 years ago

The gravitational attraction between two objects increases if

Physics

1 answer:

JulijaS [17]3 years ago

8 0

Answer:

they are closer!!

Explanation:

Hope this helped!! :D

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A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

$f_r = 150.47 N$

8 0

3 years ago

The diameter of a copper atom is approximately 2.28e-10 m. The massof one mole of copper is 64 grams. Assume that the atoms area

KIM [24]

1) Mass of one copper atom: $1.063\cdot 10^{-22} kg$

2) There are $9.33\cdot 10^{24}$ atoms in the cube

3) Mass of the cubical block: 992 kg

Explanation:

We are told here that the mass of one mole of copper is

$M=64 g$

for

$n=1 mol$ (number of moles)

We also know that the number of atoms inside 1 mole of substance is equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

This means that $N_A$ atoms of copper have a mass of M = 64 g. Therefore, we can find the mass of one copper atom by dividing the total mass by the avogadro number:

$m=\frac{M}{N_A}=\frac{64}{6.022\cdot 10^{23}}=1.063\cdot 10^{-22} kg$

We are told that the diameter of a copper atom is

$d=2.28\cdot 10^{-10} m$

We can assume that the atoms are arranged in a cube, and that they are all attached to each other; so the side of the cube can be written as size of one atom multiplied by the number of atom per side:

$L=Nd$

where

N is the number of atoms (rows) in one side of the cube

Since the side of the cube is

L = 4.8 cm = 0.048 m

We find N:

$N=\frac{L}{d}=\frac{0.048}{2.28\cdot 10^{-10}}=2.11\cdot 10^8$

This is the number of atom rows per side; therefore, the total number of atoms in the cube is

$N^3=(2.11\cdot 10^8)^3=9.33\cdot 10^{24}$

The total mass of the cubical block of copper will be given by the mass of one atom of copper multiplied by the total number of atoms, so:

$M= N^3 m$

where:

$N^3 = 9.33\cdot 10^{24}$ is the number of atoms in the cube

$m=1.063\cdot 10^{-22} kg$ is the mass of one atom

Therefore, substituting, we find:

$M=(9.33\cdot 10^{24})(1.063\cdot 10^{-22})=992 kg$

So, the mass of the cubical block is 992 kg.

Learn more about mass and density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

3 0

2 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

I need HELP Please !!

aliya0001 [1]

You have to figure it out

5 0

2 years ago

The magnetic field produced by a long straight current-carrying wire is

alexdok [17]

Answer:

proportional to the current in the wire and inversely proportional to the distance from the wire.

Explanation:

The magnetic field produced by a long, straight current-carrying wire is given by:

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

where

$\mu_0$ is the vacuum permeability

I is the current intensity in the wire

r is the distance from the wire

From the formula, we notice that:

- The magnitude of the magnetic field is directly proportional to I, the current

- The magnitude of the magnetic field is inversely proportional to the distance from the wire, r

Therefore, correct option is

proportional to the current in the wire and inversely proportional to the distance from the wire.

8 0

3 years ago