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Eduardwww [97]
3 years ago
10

Two Questions Worth 14 points(Gradpoint)

Physics
1 answer:
Liula [17]3 years ago
7 0
<span>1. Pu-239 has a half-life of _____.
</span>  24,100 years
2. <span>Nuclear fusion has not yet been harnessed as an energy source because _____.</span>
a.sufficient pressure cannot be generated  
You might be interested in
A bird is traveling west with a velocity of 8 m/s and momentum of 40 mg*m/s what's the birds mass?
VikaD [51]

Answer:

m = 5 [mg]

Explanation:

We must remember that the definition of linear momemtum is defined as the product of mass by distance.

P = m*v

P = momentum = 40 [mg*m/s]

m = mass [mg]

v = velocity = 8 [m/s]

Now clearing m:

m = P/v

m = 40/8

m = 5 [mg]

3 0
3 years ago
true or false/ force field that exists in the space around every massive body is called electric field​
musickatia [10]

Answer:

True

Explanation:

An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

3 0
3 years ago
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni
irinina [24]

The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

The given parameters;

  • <em>length of the solenoid, L = 91 cm = 0.91 m</em>
  • <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
  • <em>number of turns of the solenoid, N = 1300 </em>
  • <em>current in the solenoid, I = 3.6 A</em>

The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

where;

\mu_o is the permeability of frees space = 4π x 10⁻⁷ T.m/A

B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T

Thus, the magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

Learn more here:brainly.com/question/17137684

7 0
2 years ago
A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the
Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object.  We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t  = 55 s

v2 = ?

We can use a kinematic equation to solve this problem.  We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0
3 years ago
For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
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