All categories

3 years ago

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

1 answer:

6 0

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

You might be interested in

A 2kg bowling ball rolls at a speed of 5 m/s on a roof of the building that is 40 meters tall. What is the kinetic energy

MrRa [10]

The kinetic energy is $\frac 1 2 m v^2$ and the height of the building doesn't matter at all.

$E = \frac 1 2 m v^2 = \frac 1 2 (2)(5)^2 = 25$ joules

8 0

3 years ago

In the above lightwave the property labeled a determines which characteristic of visible light

kap26 [50]

The light must be either very dim or else non-existent.
We can't see the light wave or the label.

7 0

3 years ago

2. Which of the following is NOT true of work?

salantis [7]

Answer:

Explanation:

Work is not a vector but it is a scalar

3 0

3 years ago

Read 2 more answers

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

4 years ago

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

8 0

3 years ago