Ask question

Ask question

All categories

7 days ago

15

An engine performs 2700 J of work on a scooter. The scooter and rider have a combined mass of 150 kg. If the scooter started at

rest, what is the speed of the bike after the work is performed?

1 answer:

serg [2.5K]7 days ago

6 0

According to the principle of energy conservation, the engine's work in moving the scooter is converted into the scooter's kinetic energy, represented as:

You might be interested in

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Softa [2029]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Reservoir pressure = 10 atm

$T_1$ = Reservoir temperature = 300 K

$P_2$ = Exit pressure = 1 atm

$T_2$ = Exit temperature

$R_s$ = Specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

Assuming isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

Flow density at exit measures 2.2721 kg/m³

4 0

1 month ago

A projectile is fired from ground level with a speed of 150 m/s at an angle 30.° above the horizontal on an airless planet where

Yuliya22 [2420]

Answer:

130 m/s (to two significant figures)

Explanation:

In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.

u represents the initial projectile velocity = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

A projectile's motion can be viewed as made up of independent vertical and horizontal elements.

The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.

Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.

Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.

At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s

6 0

6 days ago

Read 2 more answers

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

Ostrovityanka [2204]

Answer: The calorimeter's heat capacity is $6.72J/g^oC$

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat capacity of water = $4.184J/g^oC$

$c_2$ = specific heat capacity of calorimeter =?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of the mixture = $35.0^oC$

$T_1$ = initial temperature of the water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now substituting all provided values into the formula, we obtain

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

$c_2=6.72J/g^oC$

Hence, the heat capacity of the calorimeter is $6.72J/g^oC$

3 0

1 month ago

A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

Ostrovityanka [2204]

Response:

$\text{heat loss} = 24864.05 \ W/m^2$

Clarification:

If

$T_1$ , $T_2$ represent the temperatures of gases and liquids in Kelvins,

$t_1$ and $t_2$ denote the thicknesses of the gas layer and steel slab in meters,
$h_1$ , $h_2$ are the convection coefficients for gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ represents the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ signify thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

by employing known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Utilizing the rate equation:

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature is $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Correspondingly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature profile is depicted in the image provided

3 0

14 days ago

The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

ValentinkaMS [2425]

Answer:3.87*10^-4

Explanation:

To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.

From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.

Solving gives frequency F=0.87*10^20 Hz.

Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.

Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).

This energy is then converted from joules to MeV.

Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding

3.87*10^-4 u.

6 0

4 days ago