What is the acceleration of the block down the incline if the angle of the incline is 38 ◦ ? Answer in units of m/s 2 .

The SAF operates the M113 Ultra APC.

HACTEHA [7]

The volume of the object must be no larger than $11.15 m^3$ .

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

$\rho = 1000 kg/m^3$

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

$\rho = \frac{m}{V}$

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

$m=1.115\cdot 10^4 kg$

Therefore, we can re-arrange the equation to find its volume:

$V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3$

So, the volume of the object must be no larger than $11.15 m^3$ .

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8 0

4 years ago

Which data set has the smallest standard deviation?

hjlf

Answer:

B. 22,22,23,23,22,22,23

Explanation:

The standard deviation is a measure of dispersion or variability of a data set. In order to determine the data set that has the smallest standard deviation, we shall investigate on the ranges of the data sets given. The range of a data set is simply the difference between the maximum and minimum values in a data set. A data set that has a smaller range also has a smaller standard deviation.

From the alternatives given, the data set given by alternative B has the smallest range and consequently the smallest standard deviation.

The maximum value is 23 while the minimum is 22. The range is 1.

5 0

3 years ago

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

MaRussiya [10]

Ans: Time taken by a pulse to travel from one support to the other = 0.348s

Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = $\sqrt{ \frac{T}{\mu} }$

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = $\sqrt{ \frac{150}{0.0232} }$ = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s

5 0

3 years ago

a grinding wheel is initially rotating with an angular velocity 5500 rad/srad/s when its motor is suddenly turned off. it comes

kiruha [24]

The angle through which the grinding wheel rotates in the first second = 5300 rad

Angular velocity is, the time charge at which an object rotates, or revolves, about an axis, or at which the angular displacement between our bodies changes. within the discern, this displacement is represented via the angle θ among a line on one body and a line on the alternative.

The angular velocity is described as the charge of trade of the angular position of a rotating body. Linear speed is defined because the charge of change of displacement with respect to time whilst the item moves alongside a straight course.

Initial angular velocity of the grinding wheel = ω1 = 5500 rad/s

Final angular velocity of the grinding wheel = ω2 = 0 rad/s (Comes to rest)

Time is taken by the grinding wheel to come to rest = T = 10 sec

Angular acceleration of the grinding wheel = α

2 = ω1 + αT

0 = 5500 + α(10)

α = - 400 rad/s2

Negative as it is deceleration.

The angle through which the grinding wheel rotates in the first second = θ

Time period = T1 = 1 sec

θ = ω₁T1 + αT1²/2

θ = (5500)(1) + (-400)(1)²/2

θ = 5300 rad

The angle through which the grinding wheel rotates in the first second = 5300 rad

Learn more about angular velocity here:-brainly.com/question/6860269

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5 0

1 year ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

$\Delta x\Delta p=\frac{h}{4\pi }$

Given that the uncertainty in x is 0.1 mm.

Therefore,

$\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}$

Therefore, uncertainty in the transverse momentum of photon is $0.53\times 10^{-30}kgms^{-1}$

6 0

3 years ago