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3 years ago

Find the mass. 10 points. Will give brainliest.

Physics

2 answers:

TiliK225 [7]3 years ago

7 0

Answer:

Explanation:

Given,

Force ( f ) = 30 N

Acceleration(a) = 7.6 m/s

Now, Let's find the mass of the ball

Using the Newton's second law of motion:

We get:

$force \: = mass \: \times acceleration$

plug the value

$30 \: = m \: \times 7.6$

Use the commutative property to reorder the terms

$30 = 7.6 \: m$

Swap the sides of the equation

$7.6m = 30$

Divide both sides of the equation by 7.6

$\frac{7.6 \: m}{7.6} = \frac{30}{7.6}$

Calculate

$m = 3.94 \: kg$

Hope this helps..

Best regards!!

FinnZ [79.3K]3 years ago

4 0

Answer:

$\displaystyle \boxed{\mathrm{3.95 \: kg }}$

Explanation:

$\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}$

$\mathrm{force = 30N}$

$\mathrm{acceleration = 7.6 \: m/s^2 }$

$\mathrm{Find \: the \: mass.}$

$\mathrm{30 = m \times 7.6}$

$\displaystyle \mathrm{m =\frac{30}{7.6} }$

$\displaystyle \mathrm{m = 3.947... }$

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When the liquid line is restricted, the supply of refrigerant to the metering device is reduced. What is the effect on suction p

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Answer:

The suction pressure decreases and the superheat increases when the liquid line is restricted and the supply of refrigerant to the metering device is reduced.

Explanation:

1. The five components of refrigeration are:

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2. Now when the cooler air meets the warm gas present in the coils, it

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3. Thus, when the liquid form is at high pressure, the refrigerant then

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4. The refrigerant also absorbs the warm air present in the fridge, which

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cycle repeats in the same form.

Thus, when the liquid line is restricted and the supply of refrigerant to the metering device is reduced it causes a decrease in suction pressure and an increase in superheat.

Learn more about refrigeration here:

<u>brainly.com/question/9046279</u>

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8 0

1 year ago

A liquid of density 830 kg/m3 flows through a horizontal pipe that has a cross-sectional area of 1.20 x 10-2 m2 in region A and

kicyunya [14]

Answer:

A) volume flow rate = 0.047 m3/s

B) mass flow rate = 39.01 kg/s

Explanation:

Detailed explanation and calculation is shown in the image below

3 0

2 years ago

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

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2 years ago

True/False: Paraphrasing Einstein’s Theory of General Relativity, gravity "handprint" makes an "indentation" in the fabric of sp

Y_Kistochka [10]

B.False

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3 years ago

Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the d

uysha [10]

Answer:

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Explanation:

This problem can be solved by using Young's double-slit experiment formula:

$Y = \frac{\lambda L}{d}$

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Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

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3 0

2 years ago