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Elis [28]
3 years ago
11

Which vector should be negative?

Physics
1 answer:
Ket [755]3 years ago
7 0

Answer: the velocity of a falling object.

As the object is falling down it is experiencing a velocity that is continuously reducing each second and becomes 0 once it reaches the ground. The graph between velocity and time in this case would be a negative slope.

Thus for a object falling down the velocity considered would be negative owing to the reducing speed.

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What is the definition of the half-life of a radioactive isotope?
Illusion [34]

Answer:Half-life is the amount of time it takes for the initial mass of the isotope to decompose, by half, into other lighter atoms.

Explanation:Different radioactive isotopes have different half-lives. For example, the element technetium-99m has a half life of 6 hours. This means that is 100 kg of the element is left to decay, in 6 hours, 50kg of the mass will have changed into other elements/atoms. The half-life of uranium-238 is 4.5 billion years while that of polonium-216 is only 0.145 seconds.

5 0
2 years ago
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A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
What of the following is the mathematical form of boyles law?
viva [34]

Answer:

C

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V=1/p

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denis23 [38]
It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open
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3 years ago
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Tresset [83]
Answer would be 5.0x104m/s2
8 0
3 years ago
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