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3 years ago

Which of the following is an example of an allocation decision?|

Physics

2 answers:

Charra [1.4K]3 years ago

8 0

Answer:

D. Coal

Explanation:

Oksana_A [137]3 years ago

7 0

Answer:

The answer is D- Coal is set aside to burn as heating fuel

Explanation:

Allocation decisions set aside resources for a specific purpose.

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A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s

Taya2010 [7]

42.47 meters per second

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3 years ago

A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t

nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

$\Phi = BA Cos \theta$

Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

$\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )$

At time $t = 5.71s$ , Induced emf is,

$\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))$

$\epsilon = 10.9V$

Therefore the magnitude of the induced emf is 10.9V

4 0

3 years ago

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Which of the following is NOT a type of acceleration?

otez555 [7]

It’s C because just trust

5 0

3 years ago

A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete

lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ $F_x=mgSin \theta=ma$

⇒ $a_x=gSin \theta$

$=9.8 \ Sin35^{\circ}$

$=5.62 \ m/s^2$

(b)

If friction present,

⇒ $F_x=mgCos \theta$

⇒ $F_x=mgSin \theta-\mu_r mgCos \theta$

⇒ $a=gSin \theta-\mu_rgCos \theta$

$=9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}$

$=5.46 \ m/s^2$

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3 years ago

What is a main difference between Mars and Saturn?

tiny-mole [99]

Answer:

Explanation:

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