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ozzi
16 days ago
15

A Top Fuel Dragster initially at rest undergoes an average acceleration of 39 m/s2 for 3.97 seconds. How far will it go in this

time?
Physics
1 answer:
serg [2.5K]16 days ago
5 0

Answer:

s = 307.34 m

Explanation:

To determine the distance the dragster traveled during the specified time, we will apply the second equation of motion, expressed as follows:

s = Vi t + (0.5)at²

where,

s = distance traveled by the dragster =?

Vi = Initial Velocity = 0 m/s

t = time interval = 3.97 s

a = acceleration = 39 m/s²

Thus,

s = (0 m/s)(3.97 s) + (0.5)(39 m/s²)(3.97 s)²

s = 307.34 m

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Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [2210]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

4 0
25 days ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
ValentinkaMS [2425]

Answer: total angular distance = 1700° and 29.7 rad

   the total angular displacement = 0

Explanation:

This is a breakdown of the calculations needed.

The task is to determine both the total angular distance and displacement experienced by the knee.

To calculate the distance traveled by the knee, consider that when squatting, the knee bends 85° to lower, and then another 85° to return to standing (upright). Thus, the cumulative angular movement during the squat totals 170°.

For 10 squats, the knee must undergo 170° motion multiplied by 10, resulting in:

10 * 170° = 1700°

As such, the total angular distance reached is 1700°.

Now converting this to radians since both degree and radian outputs are required:

Since 2π equals 360°, it follows that one degree equates to about 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1700° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

For the second part, remember that angular displacement is determined by the angular distance divided by time, leading to a displacement of zero because the knee's ending position is the same as its starting position.

I hope this is helpful!!!!

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5 0
18 days 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 [2208]

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

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14 days ago
. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be
Softa [2035]
A 40 kg child throws a 0.5 kg stone at a velocity of 5 m/s. To find the recoil, we apply the conservation of momentum formula: m1•v1 + m2•v2 = 0, where m1 is the mass of the child, and v1 is the child's recoil velocity. Applying the known values results in 40•v1 = -0.5 × 5, leading to v1 = -2.5 / 40, which simplifies to v1 = -0.0625 m/s. Thus, the child's recoil speed is 0.0625 m/s.
6 0
16 hours ago
The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block
serg [2593]

Answer: 592.37m

Explanation:

Person D is represented by the blue line.

The total displacement is calculated by subtracting the initial position from the final position. Starting at (0,0), the path consists of moving down two blocks, then right six blocks, followed by moving up four blocks, and finally left one block.

Here, I consider the positive direction of the x-axis to the right and the positive direction of the y-axis as upward.

Thus, the new coordinates will be, with B representing a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

Given that B = 110m

P = (550m, 220m)

The displacement corresponds to the length of the vector, since the change from the initial position (0,0) to P is simply P:

P = √(550^2 + 220^2) = 592.37m

4 0
1 month ago
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