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earnstyle [38]
3 years ago
7

The is greater for low-mass stars than it is for high-mass stars. 2. The stars known as are the very largest and brightest of al

l the stars. 3. Most of the stars near the Sun are . 4. Stars are classified on the basis of their spectral type and 5. Cepheids are examples of . 6. are no longer generating energy through nuclear fusion. 7. Stars that are cooler than the Sun yet 100 to 1,000 times as luminous as the Sun are classified as .
Physics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

red giant stars/ red hyper giants

Explanation:

take stephenson 2-18 for example the star is only 3200k where the sun is around 5000k.

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
antoniya [11.8K]

Answer:

r=5.278\times 10^{-4}\ m

Explanation:

Given that:

  • magnetic field intensity, B=0.07\ T
  • kinetic energy of electron, KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J
  • we have mass of electron, m=9.1\times 10^{-31}\ kg

<em>Now, form the mathematical expression of Kinetic Energy:</em>

KE= \frac{1}{2} m.v^2

1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2

v^2=4.2198\times 10^{11}

v=6.496\times 10^6\ m.s^{-1}

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

r=\frac{m.v}{q.B}

r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}

r=5.278\times 10^{-4}\ m

6 0
3 years ago
If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,
Sladkaya [172]

Answer:

<h3>1.01 s</h3>

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

5 = 4.905t²

t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

6 0
3 years ago
Should genetic genealogy be used to solve crimes ? Yes or no and why?
zheka24 [161]

Answer:

yes

Explanation:

because it will help find who did the crime and it can also open jobs and opportunities for the people who likes sicnenae

8 0
3 years ago
Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
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