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Tanzania [10]
3 years ago
13

Two objects, one having three times the mass of the other, are dropped from the same height in a vacuum. At the end of their fal

l, their velocities are equal because the force of gravity is the same for both objects. Is this not true?
Physics
2 answers:
Feliz [49]3 years ago
7 0

Answer:

For two or more bodies of different mass released from height in a vacuum have the same velocity but varying force

Explanation:

Consider a body H with initial velocity (u) and final velocity V undergoing acceleration a and covering a distance( s)

From Network equation of motion it can be seen that

V^2=u^2+2as

From this it can be seen that velocity is not dependent on the the masses of the body.

Rather it depends on acceleration due to gravity which is a constant for both of the body

gogolik [260]3 years ago
5 0

Answer:

Two objects will have the equal velocities but the forces on both of them will not be equal. The equal velocities of these objects are due to their equal acceleration.

Explanation:

From the newton's equation

v^{2} -u^{2} = 2as

so here we can say that velocity does not depends on the mass.

The acceleration of both objects will be same but not the forces because

F = Ma

As the force is depending on the mass so it will not be the same for both objects.

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Rachel has an unknown sample of a radioisotope listed in the table. Using a special technique, she is able to measure the mass o
mart [117]
<span>Barium-139 is the correct answer.</span>
3 0
3 years ago
Read 2 more answers
The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt
drek231 [11]

Answer:

5 mm

Explanation:

Youngs's modulus (Y) is described by the following expression:

Y=\frac{F*L}{\Delta L*A}

Where F is the force exerted on the tendon, L is its length, A is its area and ΔL is its change in length (stretching).

The force in this case is 8 times the weight of the runner:

F= 8*m*g\\F= 8*70*9.8\\F=5488 N

Therefore, the change in length of the tendon is given by:

\Delta L=\frac{F*L}{Y*A}\\\Delta L=\frac{5488*0.15}{0.15*10^{10}*1.1*10^{-4}}\\\Delta L= 0.004989 m

the runner's Achilles tendon will stretch by 0.004989 m, which is roughly 5 mm.

8 0
3 years ago
An open organ pipe is 2.46 m long, and the speed of the air in the pipe is 345 m/s.
lukranit [14]

Answer:

Fundamental frequency is 70.12 m

Explanation:

For an open organ pipe, the fundamental frequency is given by :

f=\dfrac{nv}{2l}

n = 1 for fundamental frequency

v is speed of sound in air, v = 345 m/s

l is length of open organ pipe, l = 2.46 m

Substituting values in above formula. So,

f=\dfrac{1\times 345}{2\times 2.46}\\\\f=70.12\ Hz

So, the fundamental frequency of this pipe is 70.12 m.

6 0
3 years ago
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho
cupoosta [38]

Answer:

xf = 5.68 × 10³ m  

yf = 8.57 × 10³ m  

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

ax = 0                          ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t²     ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

yf = 8.57 × 10³ m  

5 0
4 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

where

\phi = Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

\lambda = wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]

Magnetic field Equation

B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

Where B₀= E₀/c

B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T

B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]

6 0
3 years ago
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