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PtichkaEL [24]
3 years ago
14

Carlos is playing darts. He throws a dart at the bullseye, but it hits the outer ring of the dartboard instead. He aims for the

bullseye again and hits it with a second dart. Carlos threw three more darts which landed close to his first shot. What does the second throw indicate about his accuracy and precision? The second throw shows poor accuracy but high precision. The second throw shows accuracy but poor precision. The second throw shows both accuracy and precision. The second throw shows poor accuracy and poor precision.
Physics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

The second throw shows both accuracy and precision.

Explanation:

The second throw shows both accuracy and precision because he knew in which degree and the amount of force that can be applied to the dart so that it hits the center of the bullseye or close to it. Due to knowing of these two things the Carlos is able to hit the dart close to the target at the second time so in this way he got both accuracy and precision at the second turn of throwing darts.

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Two astronauts of mass 100 kg are 2 m apart in outer space. What is the
fredd [130]

The force of gravity between the astronauts is 1.67\cdot 10^{-7}N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

m_1 = 100 kg\\m_2 = 100 kg

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N

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4 0
3 years ago
Read 2 more answers
ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha
Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0
3 years ago
If I bought 59 kids off the black market in a secret door in the stalls and I sell 21 kids then ask for 81 more then 12 off the
larisa [96]

You will be left with 106 kids

<h3>Meaning of word problem</h3>

A word problem can be defined as a mathematical problem that is written in word or written in a sentence format.

In a word problem, the student is expected to decode the sentence into a mathematical expression before solving

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5 0
2 years ago
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0
3 years ago
A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu
Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

                                                     = 25° C

The specific heat capacity of aluminium, c = 900  J/kg°C

The formula for thermal energy,

                             <em>Q = mcΔT</em>

                                 = 1.0 x 900 x 25

                                 = 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0
3 years ago
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