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tatiyna
3 years ago
13

Which of the following is true of all minerals?

Physics
2 answers:
vodka [1.7K]3 years ago
6 0

I think that the answer D. Made of one or more elements from the periodic table

9966 [12]3 years ago
5 0

Answer:

D

Explanation:

I got it correct on the test

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5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈  4513.093·d  

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0
3 years ago
how does a sprinter sprint what is the forward force on a sprinter as she accelerates. where does that force come from
NARA [144]

On the starting blocks, sprinters use their feet to push backward. The blocks respond by pressing forward with a force equal to this with their feet.

<h3>What drives the sprinter forward?</h3>

Vertical forces must be larger than the pull of gravity in order to assist the sprinter in moving forward as gravity is pushing him or her downward. The propulsive force is the force that propelled the runner forward.

<h3>Basketball players must jump straight up into the air, but how?</h3>

An interaction diagram and a free-body diagram should be included in your explanation. The player pushes down on the ground, which pushes up against him in return. As a result of his push being stronger than gravity, the player accelerates upward.

To know more about sprinter forward visit:-

brainly.com/question/14310782

#SPJ4

8 0
1 year ago
5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

jack JACKKK

Explanation:

OK YAHHHHH SO DO THIS

3 0
4 years ago
A square coil of wire of side 3.95 cm is placed in a uniform magnetic field of magnitude 2.25 T directed into the page as in the
Len [333]

Answer:

Explanation:

In this case we shall calculate rate of change of flux in the coli to calculate induced emf .

Flux through the coil  = no of turns x area x magnetic field perpendicular to it

=34 x  2.25 x (3.95 )²x 10⁻⁴ Weber

= 1193.4  x 10⁻⁴Weber

Final flux through the coil after turn by 90°

= 1193.4 x 10⁻⁴ cos 90 ° =0

Change of flux

= 1193.4 x 10⁻⁴ weber.

Time taken = 0.335 s .

Average emf= Rate of change of flux

= change in flux / time

=1193.4 x 10⁻⁴ / .335

= 3562.4 x 10⁻⁴

356.24 x 10⁻³

=356.24 mV.

Current induced = emf induced / resistance

= 356.24/.780

= 456.71 mA.

8 0
3 years ago
The resultant of two forces is 250 N and the same are inclined at 30° and 45° with resultant one on either side calculate the ma
Varvara68 [4.7K]

Answer:

The two forces are;

1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force

2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force

Explanation:

The given parameters are;

The (magnitude) of the resultant of two forces = 250 N

The angle of inclination of the two forces to the resultant = 30° and 45°

Let, F₁ and F₂ represent the two forces, we have;

F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force

The components of F₁ are \underset{F_1}{\rightarrow} = -F₁ × sin(30°)·i + F₁ × cos(30°)·j

The components of F₂ are \underset{F_2}{\rightarrow} = F₂ × sin(45°)·i + F₂ × cos(45°)·j

The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j

The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out

We have;

F₂ × sin(45°)·i = F₁ × sin(30°)·i

F₂ ·√2/2 = F₁/2

∴ F₁ = F₂ ·√2

∴ F₂ × cos(45°)·j  + F₁ × cos(30°)·j = 250·j

Which gives;

F₂ × cos(45°)·j  + F₂ ·√2 × cos(30°)·j = 250·j

F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N

F₂ ≈ 129.41 N

F₁  = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N

F₁  ≈ 183.013 N

The two forces are;

A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.

5 0
3 years ago
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