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BlackZzzverrR [31]
2 years ago
10

Answer part (d) please

Physics
2 answers:
attashe74 [19]2 years ago
7 0

Average speed = (distance covered) / (time to cover the distance)

For the full 18 seconds described by the graph . . .

Average speed = (16 meters) / (18 seconds)

Average speed = (16 / 18) m/s

<em>Average speed =  0.89 m/s</em>

finlep [7]2 years ago
3 0

Answer:

MARK me brainliest please and follow my page

Explanation:

All you have to do to get the average speed is to calculate the total distance covered and divide it by the total time taken

= 16/18 = 0.88m/s

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Which of the following is not a colligative property? a. osmotic pressure b. lattice energy c. vapor-pressure lowering d. boilin
Gre4nikov [31]

Answer:

b) lattice energy

Explanation:

A solution is said to have colligative property when the property depends on the solute present in the solution.

Colligative property depend upon on the solute particle or the ion concentration not on the identity of solute.

osmotic pressure, vapor pressure lowering , boiling point elevation and freezing point lowering all depend upon solute concentration so they will not have colligative property so, the answer remains option 'b' which is lattice energy.

8 0
3 years ago
Find the tension in the two ropes that are holding the 4.2 kg object in place.
Blababa [14]

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

7 0
3 years ago
The graph shows position versus time for an object moving with the constant acceleration. what is the final velocity of the obje
natali 33 [55]

Answer:

  3.0 m/s

Explanation:

The equation of motion for constant acceleration (a) is ...

  x(t)=\dfrac{1}{2}at^2 +v_0t+x_0

The problem statement tells us v₀ = 1, and we read from the graph that x₀=1. We also read from the graph that x(10) = 21. Filling these values into the equation, we can find <em>a</em> and x'(10).

  21 = \dfrac{1}{2}a(10^2)+1(10)+1\\\\10=50a\qquad\text{subtract 11}\\\\a=\dfrac{1}{5}\\\\x'(t)=at+v_0\\\\x'(10)=\dfrac{1}{5}\cdot 10+1=3.0

The final velocity of the object at t=10 s is about 3.0 m/s.

_____

<em>Comment on the graph</em>

We note this graph better represents increasing acceleration than it does constant acceleration. x(2) = 3.4 per the equation. It is graphed as about 4.

8 0
3 years ago
A radiograph of the forearm is produced using 4 mA at 75 kVp. What kvp would be required to double the exposure?
BabaBlast [244]

Answer:

Required KVP = 86 KVP (Approx)

Explanation:

Given:

Current KVP = 75 KVP

Find:

KVP required to double exposure

Computation:

According to 15% rule of KVP,

15% change increse in KVP required to get double exposure.

So,

Required KVP = Current KVP + Current KVP(15%)

Required KVP = 75 KVP + 75 (15%)

Required KVP = 75 KVP + 11.25 KVP

Required KVP = 86.25 KVP

Required KVP = 86 KVP (Approx)

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3 years ago
Una esfera homogénea de radio r y peso W resbala en un piso bajo la acción de una fuerza horizontal constante P aplicada a una c
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