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Elina [12.6K]
3 years ago
9

two objects are connected by a light string that passes over a frictionless pulley as in the figure below, where m1 A) a1=g

B) a1>g
C) a1 D) a1>a2
E) a1
Physics
1 answer:
Shtirlitz [24]3 years ago
5 0
You didn't attach the figure. Your text is incomplete. And you never got around to asking a question. Other than that, we're on it.
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How to answer this question?
laila [671]

Answer:

measure the vector diagram first

5 0
2 years ago
How to find acceleration without final velocity?
Andrews [41]
Acceleration is the rate of change of a the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. However, if we are not given the final velocity, it would not be possible to use the kinematic equations. One possible to calculate this value would be by generating an equation of distance with respect to time and getting the second derivative of the equation.
7 0
3 years ago
The altitude of a hang glider is increasing at a rate of 6.75 m/s. At the same time, the shadow of the glider moves along the gr
Rufina [12.5K]

Answer:

16.45 m/s

Explanation:

Let y be the vertical distance and x be the horizontal distance

We are given that

The altitude of hang glider increasing at the rate=v_y=\frac{dy}{dt}==6.75m/s

The shadow of the glider moves along the ground at speed=v_x=\frac{dx}{dt}=15m/s

We have to find the magnitude of glider's velocity.

We know that

Magnitude of velocity=v=\sqrt{v^2_x+v^2_y}

Substitute the values

v=\sqrt{(15)^2+(6.75)^2}

v=\sqrt{225+45.5625}

v=16.45m/s

Hence, the magnitude of glider's velocity=16.45 m/s

4 0
3 years ago
I really need help with have no clue
nexus9112 [7]
I would say C and A.

It’s the only option that’s the correct length apart.
8 0
3 years ago
old stars obtain part of their energy by the fusion of three alpha particles to form a 12/6c nucleus, whose mass is 12.0000 u. H
marta [7]

Answer:

11.6532 x 10⁻¹¹ J or 7.3 MeV is given off

Explanation:

Mass of an alpha particle = 4.0026u,   ∴ mass of three = 12.0078u

Find the difference in mass.

Mass of three alpha - Mass of Carbon nucleus

12.0078u - 12u = 0.0078u

Since 1u = 1.66 x 10⁻²⁷ kg

Therefore, 0.0078u = 1.2948 x 10⁻²⁷

Now that we know Mass(m) = 1.2948 x 10⁻²⁷ and Speed (c) 3 x 10⁸ m²s⁻²

Formular for Energy ==> E₀ = mc²

E = (1.2948 x 10⁻²⁷) (3 x 10⁸ m²s⁻²)²

E = (1.2948 x 10⁻²⁷) (9 x 10¹⁶) J

E = 11.6532 x 10⁻¹¹ J

Or, if you need your energy in MeV

1 MeV = 1.60x10⁻¹³ J

Just do the conversion by dividing 11.6532 x 10⁻¹¹ J by 1.60x10⁻¹³ J

It will give you 7.3 MeV

7 0
3 years ago
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