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3 years ago

Moving at 3.2m/s slows down after 32 seconds moving at 1.2m/s What is his acceleration?

Physics

1 answer:

Stels [109]3 years ago

8 0

Answer:

His acceleration is -0.0625 m/s².

Explanation:

Given:

Initial velocity is, $u=3.2\ m/s$

Final velocity is, $v=1.2\ m/s$

Time taken for the change is, $t=32\ s$

Acceleration of a body is defined as the rate at which the velocity is changing. Acceleration is positive if there is increase in velocity with time and acceleration is negative when the body slows down or there is decrease in velocity with time.

Here, there is decrease in velocity over the given time. Therefore, the acceleration is negative and is given as:

$a=\frac{v-u}{t}\\\\a=\frac{1.2-3.2}{32}\\\\a=\frac{-2}{32}=-0.0625\ m/s^2$

Therefore, his acceleration is -0.0625 m/s².

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If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
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3 years ago

Read 2 more answers

Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.

-BARSIC- [3]

The total resistance in a series circuit is equal to the sum of all resistors (R total = ΣRi).

R total = R1 + R2 + R3 = (3 + 4 + 5) Ω = 12 Ω

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3 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:

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

5 0

3 years ago

Newtown third law applies to blank of objects

fomenos

Answer:

All

Explanation:

I'm not sure what you meant but Newton's third law which basically states that every action has an equal and opposite reaction applies to all objects. So I think the answer is all.

8 0

3 years ago

Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet

Vikki [24]

Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

$Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\$

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

$340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\$

v₂ = 5.7 m/s

6 0

3 years ago