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Marina CMI [18]
2 years ago
12

Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an

Physics
2 answers:
dlinn [17]2 years ago
8 0

Answer:

\boxed {\boxed {\sf C. \ 23.33 \ kg}}

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F=ma

The mass of the ball is unknown. The ball is accelerating at 1.8 meters per second squared. An unbalanced force of 42 Newtons is applied to the ball.

Convert the units of force. 1 Newton is equal to 1 kilogram meter per second squared, so our answer of 42 Newtons is equal to 42 kg*m/s².

  • F= 42 kg*m/s²
  • a= 1.8 m/s²

Substitute the values into the formula.

42 \ kg*m/s^1 = m * 1.8 \ m/s^2

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 1.8 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 1.8 m/s².

\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} = \frac{a*1.8 \ m/s^2}{1.8 \ m/s^2}

\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} =m

The units of meters per second squared cancel.

\frac {42 \ kg}{1.8 } =m

23.3333333 \ kg=m

Round to the hundredth place. The 3 in the thousandth place tells us to leave the 3 in the hundredth place.

23.33 \ kg \approx m

The mass of the ball is approximately <u>23.33 kilograms.</u>

Juli2301 [7.4K]2 years ago
4 0

Answer:

<h2>23.33 kg </h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m  = \frac{42}{1.8}   = 23.3333... \\

We have the final answer as

<h3>23.33 kg</h3>

Hope this helps you

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Dovator [93]

Answer:

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

Explanation:

initial veetical speed V₀y=0

Horizontal speed Vx = Vx₀= 3.80m/s

Vertical drop height= 3.90m

Let Vy = vertical speed when it got to the water downward.

g= 9.81m/s² = acceleration due to gravity

From kinematics equation of motion for vertical drop

Vy²= V₀y² +2 gh

Vy²= 0 + ( 2× 9.8 × 3.90)

Vy= √76.518

Vy=8.747457

Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below

V= √Vy² + Vx²

V=√3.80² + 8.747457²

V=9.537m/s

The angle can also be calculated as

θ=tan⁻¹(Vy/Vx)

tan⁻¹( 8.747457/3.80)

=66.52⁰

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

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A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th
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The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy \frac{1}{2}mv^2

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force \frac{mv^2}{r}.

So,

Vq = \frac{1}{2}mv^2

and

Bqv = \frac{mv^2}{r}

Right from these eliminating v, we can derive

m = \frac{B^2r^2q}{2V}

On substituting the value, we get:

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answer:

thank youu for the points cute drawing by the way (✿◠‿◠)

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3 years ago
Read 2 more answers
(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t
Alex73 [517]

Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

Q = m * Lf

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Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

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Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.
Mars2501 [29]

Answer:

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We have given mass of the air particle m=5\times 10^{-16}kg

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Temperature is given T = 27^{\circ}C=273+27=300K

We have to find the root mean square speed of the particle

Which is given by v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec

So rms speed of the particle will be 38.68\times 10^8m/sec

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3 years ago
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