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3 years ago

11

A white-blue star is hotter than a red star.

2 answers:

Inessa05 [86]3 years ago

5 0

Blue light is a higher frequency than red light. This means that things that glow red are cooler than things that glow blue. So the answer is “TRUE” because white-blue are hotter than red star so red stars are cooler!

Katena32 [7]3 years ago

4 0

This is true!!

Good luck hope this helped!

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"A hiker starts at point P and walks 2.0 kilometers due east and then 1.4 kilometers due north. The vectors in the diagram below

Law Incorporation [45]

Here, you need to use your "Protractor" as it is given in the question, but we can calculate the value with the help of our mathematical calculation too:
[ Protractor can be use only in real life, not here ]

Draw an imaginary line from initial position to final position.
Now, In that triangle, tan x = P/B
tan x = 1.4 / 2
tan x = 0.70
x = tan⁻¹ (0.70)
x = 35 [ tan 35 = 0.70 ]

In short, Your Answer would be 35 degrees

Hope this helps!

3 0

3 years ago

Momentum quiz for physics

jek_recluse [69]

Answer:

p=mv=(813kg)(17)= 13,821 kg m\s

7 0

3 years ago

A classroom that normally contains 40 people is to be airconditioned with window air-conditioning units of 5-kW cooling capacity

oksano4ka [1.4K]

Answer:

About 2 units.

Explanation:

We assume that there is no heat dissipating instrument in the room

Total cooling load of the room is defined from the given below equation

$Q_{cooling}$ = $Q_{light}$ + $Q_{people}$ + $Q_{heat gain}$

where

$Q_{light}$ = 10*100 W =1 KW

$Q_{people}$ = 40*360 KJ/h= 4 KW

$Q_{heat gain}$ =15000KJ/h= 4.17 KW

$Q_{cooling}$ = 1+4+4.17=9.17 KW

the number of air conditioner unit is = $\frac{9.17}{5}$ =1.83

which is approximately 2 units

7 0

3 years ago

A 11kg block slides up a 30° inclined plane at a constant velocity. The coefficient of friction between the block and the plane

astra-53 [7]

Answer:

The magnitude of the applied force is 94.74 N

Explanation:

Mass of the block, m = 11 kg

Angle of inclination of the plane, $\theta = 30^{\circ}$

Friction coefficient, $\mu_{k} = 0.2$

Now,

Normal force that acts on the block is given by:

$F_{N} = mgcos\theta + Fsin\theta$ (1)

Now, to maintain the equilibrium parallel to ramp the forces must be balanced.

Thus

$Fcos\theta = \mu_{k}F_{N}$ (2)

From eqn (1) and (2)

$Fcos\theta = \mu_{k}(mgcos\theta + Fsin\theta)$

$F(cos\theta - \mu_{k}sin\theta) = \mu_{k}mgcos\theta$

$F = \frac{\mu_{k}mgcos\theta}{cos\theta - \mu_{k}sin\theta}$

$F = \frac{0.2\times 11\times 9.8cos30^{\circ}}{cos30^{\circ} - 0.2\times sin30^{\circ}}$

F = 94.74 N

3 0

3 years ago

A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted

Mazyrski [523]

Answer:

Mu,k between the floor and the box is 0.609

Explanation:

given information:

weight of the book, W = 325 N

applied force, F = 425 N

angle, θ = 35.2°

to find the coefficient kinetic between the floor and the box, wee need to calculate the horizontal and vertical force.

first we calculate the vertical force, there are weight, normal force(N) and applied vertical force.

according to Newton's first law

ΣF = 0

Σ $F_{y}$ = 0

W - N + F sin θ = 0

N = W + F sin θ

Next, we calculate the force in horizontal direction. applied force and friction force( $F_{friction}$ )

Σ $F_{x}$ = 0

F cos θ - $F_{friction}$ = 0

F cos θ - μk (W - F sin θ) = 0

μk = (F cos θ) / (W + F sin θ)

= (425 cos 35.2°)/(325+425 sin 35.2°)

= 0.609

3 0

3 years ago