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Rus_ich [418]
3 years ago
12

Timed please hurry

Physics
2 answers:
vladimir1956 [14]3 years ago
5 0
The answer is D. disorganized
Naddika [18.5K]3 years ago
5 0

Answer:

the answer is disorganized I think

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A 0.2-stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone
Nikitich [7]

Answer:

2960 N

Explanation:

Convert rev/min to rad/s:

150 rev/min × (2π rad/rev) × (1 min / 60 s) = 50π rad/s

Sum of forces in the centripetal direction:

∑F = ma

T = m v² / r

T = m ω² r

T = (0.2 kg) (50π rad/s)² (0.6 m)

T = 2960 N

3 0
3 years ago
A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3
Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0
3 years ago
What is the target heart rate for a 24 year old?
andriy [413]

Answer:

196

Explanation:

subtract 24 from 220 to get your answer.

6 0
3 years ago
Read 2 more answers
I have a bottle of gas, the bottle can expand and contract. Initially the gas is at 1 kpa of pressure and a volume of 1 Liter, a
drek231 [11]

Answer:

P₂ = 1.22 kPa

Explanation:

This problem can be solved using the equation of state:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

where,

P₁ = initial pressure = 1 KPa

P₂ = final pressure = ?

V₁ = initial Volume = 1 liter

V₂ = final volume = 1.1 liter

T₁ = initial temperature = 290 k

T₂ = final temperature = 390 k

Therefore,

\frac{(1\ kPa)(1\ liter)}{290\ k} =\frac{(P_2)(1.1\ liter)}{390\ k}\\\\P_2= \frac{(1\ kPa)(1\ liter)(390\ k)}{(290\ k)(1.1\ liter)}

<u>P₂ = 1.22 kPa</u>

7 0
3 years ago
Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o
Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

         qA = ½ q

         qB = ½ q

         qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

         qA = 1/2 (q / 2) = ¼ q

         qC = ¼ q

         qB = ½ q

At this point all spheres have a charge,

      qA = ¼ q

      qb = ½ q

      qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

       Q = ½ q + ¼ q = ¾ q

When splitting the charge

        qB = ½ ¾ q = 3/8 q

        qC = ½ ¾ q = 3/8 q

        qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

       Q = qA + QB + QC

       Q = ½ q + 3/8 q + ¼ q

        Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0
3 years ago
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