Ask question

Ask question

All categories

Anestetic [448]

3 years ago

12

A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

ng makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant

1 answer:

enyata [817]3 years ago

7 0

Answer:

$T=8.1N$

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

$\alpha=90-\theta\\\\\alpha=90-40$

$\alpha=50 \textdegree$

Generally the equation for Tension is is mathematically given by

$T=\frac{mv^2}{r}+mgcos \alpha$

$T=\frac{0.40*5^2}{1.8}+0.40*5cos50$

$T=8.1N$

You might be interested in

A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t

Murljashka [212]

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

8 0

3 years ago

A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor

Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is located on the x-axis= $\lambda_1$

Linear charge density of another long line which is located on the y-axis= $\lambda_2$

We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

The direction of electric field at z=a on the positive z-axis is positive z-direction .

Because $\lambda_1$ and $\lambda_2$ are positive and the electric field is applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

6 0

3 years ago

77julia77 [94]

Answer:

What is the best description for the volume of air volume of air provided in a high quality rescue breath?

Explanation:

A) only Enough air to create a visible rise of the chest.

B) Until you can no longer force air in.

C) plenty of air make sure it is adequate to sustain life

D) clear and obvious rise of the chest, sustained over a few seconds

5 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:

1)

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:

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

2)

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

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

3)

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

3 years ago

The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle

Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

Re = ρ v D /μ

The units of each term are

ρ = [kg / m³]

v = [m / s]

D = [m]

μ = [Pa s]

The pressure

Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

Re = [kg / m³] [m / s] [m] / [kg / m s]

Re = [kg / m s] / [m s / kg]

RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity μ = 11.2 10⁻⁶ Pa s

the density ρ = 0.656 kg / m³

D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

Re = 1.19 10⁴

4 0

3 years ago