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3 years ago

3. An object of mass 90 kg travels down a slide.

Physics

1 answer:

Zolol [24]3 years ago

6 0

Gpe= 90* 9.81 * 15 = 13243.5 Joules

KE= 13243.5 =1/2 m v^2
v^2= 295.3
v= 17.2 m/s

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A particle moves along a straight line and its position at time t is given by

Alisiya [41]

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

if a(t) >0, it means 12t-48>0, and t >4, I = ]4, infinity[

the particle is speeding down

if a(t) < 0, it means 12t-48>0, and t < 4, I = ] -infinity, 4[

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

moving backward if v(t)< 0

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

so after taking care the sign of 6t² - 48 t +90,

v(t)> o when t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward

v(t) < o when t is in I = ]1, 4[,the particle is moving backward

3 0

3 years ago

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

6 0

2 years ago

Read 2 more answers

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

kherson [118]

Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

In this case we have an electron (-e) and a proton (e), so:

$F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N$

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

$F=2.64*10^{-16}N$

3 0

3 years ago

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

At State 1:

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

At State 1:

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

At State 1;

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

At State 2:

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0

2 years ago

An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi

AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

7 0

2 years ago