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3 years ago

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Object A moves with a constant velocity of -10 m/s and object B moves with a constant velocity of 5 m/s. Which object has the la

rger speed?

1 answer:

KonstantinChe [14]3 years ago

8 0

Object A has the larger speed because speed is not a vector quantity, it is scalar. This means that directionality, or in this case the sign of the velocity, doesn’t matter.
Speed = |velocity|
Object a speed =|-10|=10 m/s
Object b speed =|5|=5 m/s

Object a is the answer.

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We would have to analyze the design of an specific turbine to determine its efficiency, however it is unlikely to achieve 50% , as todays turbines have an average efficiency in the 20-35%

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3 years ago

Read 2 more answers

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

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3 years ago

I’ll give brainliest and 10 points

leonid [27]

Answer:

what's these its D okkkk

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3 years ago

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We have that the sea level pressure for Leh area is 1150mb mathematically given as

Ps= 1150 mb

<h3> Sea level pressure</h3>

Question Parameters:

Ladakh is 800 mb.

<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m

increase in height with respect to sea level corresponds to 10 mb pressure,

Generally, for 3500m the pressure change will be 350 mb.

Therefore, here for the sea level <em>pressure</em> we need to add,

Ps=800+350

Ps= 1150 mb

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2 years ago

A single-turn circular loop of wire of radius 65 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

BabaBlast [244]

Answer:

Explanation:

change in flux = no of turns x area of loop x change in magnetic field

= 1 x π 65² x 10⁻⁶ x ( 650 - 350 ) x 10⁻³

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