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gavmur [86]
3 years ago
10

Determine the magnitude of the force p required to hold the 180-kg crate in equilibrium.

Physics
1 answer:
Nadusha1986 [10]3 years ago
6 0
When tackling a problem in equilibrium, always keep in mind that all forces in any direction are all balance out. Let us imagine that this 180 kg crate is being held by a wire. The Force "p" necessary to put this crate in equilibrium will be the tension on wire acting upward. Remember that weight "W"is a force, and force according to Newton's law is equal to mass multiply by acceleration. F=W=mg. The total force needed to balance the mass is simply Force "p=mg". Therefore the magnitude of the force will be just multiplying mass and the normal pull of gravity, equivalent to 9.8m/s².  
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A wavelength is the distance between two
pentagon [3]
C. crests. The wavelength is the distance between two specific parts of a wave. This might also be the trough.
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4 years ago
On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (u) 46Ti 77.600% 45.95263
allsm [11]

Answer: Average atomic mass of titanium on that planet is 46.52

Explanation:

Mass of isotope Ti-46 = 45.95263

% abundance of isotope Ti-46  = 77.600 % = \frac{77.600}{100}=0.776

Mass of isotope Ti- 48= 47.94795

% abundance of isotope Ti-48 = 16.100%= \frac{16.100}{100}=0.161

Mass of isotope Ti- 50 = 49.94479

% abundance of isotope Ti-50  = 6.300%= \frac{6.300}{100}=0.063

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

Z=\sum[(45.95263 \times 0.776)+(47.94795 \times 0.161)+(49.94479\times 0.063)

Z=46.52

Average atomic mass of titanium on that planet is 46.52

8 0
3 years ago
A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giv
maksim [4K]

Answer:

c. 8.67 cm/s

Explanation:

From the law of conservation of momentum,

Total momentum before the thread was burned = Total momentum after was burned

mu + m'u' = mv + m'v'...................... Equation

Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.

Note: Both cart where momentarily at rest, as  such u = u' = 0. i.e the total momentum before the thread was burn = 0

And assuming the left is positive,

We can rewrite equation 1 as

mv + m'v' = 0............................................ Equation 2

Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s

Substitute into equation 2,

4.5v + 1.5(26) = 0

4.5v + 39 = 0

4.5v = -39

v = -39/4.5

v = -8.67 cm/s.

Note: v is negative because it moves to right.

Hence the velocity of the 4.5 kg cart = 8.67 cm/s.

The right option is c. 8.67 cm/s

7 0
3 years ago
Which of the following terms refers to the amount of thermal energy needed
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Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

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