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2 years ago

10

If someone were to be researching and studying similarities between our sun and a neighboring star, what kind of scientific inve

stigation would they be conducting

1 answer:

OLEGan [10]2 years ago

6 0

Answer:

a

Explanation:

because there both in the solar system

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Hii please help i’ll give brainliest!!

den301095 [7]

Answer:

C

Explanation:

6 0

2 years ago

Read 2 more answers

An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 7 m/s over

stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

= 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

t=√2(xf-xi)/a

t=3.765×10^-10 s

now we can use the power P Eq.

P=W/Δt => ΔK/Δt

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0

3 years ago

A snowball with a mass of 85 g hits the top hat of a 1.5 m tall snowman and sticks to it. the hat and the snowball, with a combi

LenaWriter [7]

Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,

Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.

v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s

Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2

Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds

Then,
x = 8*0.553 = 4.424 m

7 0

3 years ago

A crate slides down a ramp that makes a 20∘ angle with the ground. to keep the crate from sliding too fast, paige pushes back on

shutvik [7]

The solution is:
Paige's force is (somewhat) against the direction of motion: Work = F * d Where F is the force; andd is the distance
Our f is 64 N and our distance is 20 and -3.6Plugging that in our equation will give us:
= 64N * cos20º * -3.6m = -217 J

8 0

3 years ago

A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal

Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, $a=g=-9.8\ m/s^2$ (Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

$S=ut+\frac{1}{2}at^2$

Plug in the given values and solve for 'u'. This gives,

$0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s$

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0

3 years ago