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3 years ago

A car travels at an average speed of 60km/h for 15 minutes. How far does the car travel in this time?

Physics

2 answers:

jok3333 [9.3K]3 years ago

4 0

Answer:

In 15 minutes the car travels a distance of 15 km.

Explanation:

lana [24]3 years ago

4 0

Explanation:

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Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l

ziro4ka [17]

Answer:

The pressure is 6570 lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

$a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s$

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

$M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3$

then, $M_n = M_1sin \beta = 3sin(35) = 1.721$

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570 lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

$M_t_1 = M_1cos \beta = 3 cos(35) = 2.458$

$w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s$

$a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s$

$v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s$

$Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7} \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o$

6 0

3 years ago

Find the final velocity of a car that accelerates at +2 m/s2 for +4m from an

Stolb23 [73]

Answer:

Explanation:

according to third equation of motion

2as=vf²-vi²

vf²=2as+vi²

vf=√2as+vi²

vf=√2as+vi

vf=√2*2*4+3

vf=√16+3

vf=4+3=7

so final velocity is 7 m/s

5 0

3 years ago

It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica

denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is $\theta = 24.2^o$

Explanation:

From the question we are told that

The length of eire let out is $L = 250 \ m$

The emf generated is $\epsilon = 40 V$

The earth magnetic field is $B = 5.0 *10^{-5} T$

The speed of the shuttle and tether is $v = 7.80 * 10^3 \ m/s$

The emf generated is mathematically represented as

$\epsilon = L\ v\ B\ sin \ \theta$

making $\theta$ the subject of the formula

$\theta = sin ^{-1}[ \frac{\epsilon}{L * B *v} ]$

substituting values

$\theta = sin ^{-1}[ \frac{40}{250 * (5*10^{-5}) *(7.80 *10^{3})} ]$

$\theta = 24.2^o$

6 0

3 years ago

PFF URGENTE!!!! TENHO TESTE AMANHAVOMORRE

gogolik [260]

Answer:

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6 0

2 years ago

An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery su

nadya68 [22]

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

$m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}$

Where:

$m_{a}$ , $m_{t}$ - Masses of arrow and target, measured in kilograms.

$v_{a,o}$ , $v_{a,f}$ - Initial and final speeds of the arrow, measured in meters per second.

$v_{t,o}$ , $v_{t,f}$ - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

$m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})$

$v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})$

If $v_{a,o} = 2.1\,\frac{m}{s}$ , $m_{t} = 0.3\,kg$ , $m_{a} = 0.0225\,kg$ , $v_{t,o} = 2.10\,\frac{m}{s}$ and $v_{t,f} = 0\,\frac{m}{s}$ , the speed of the arrow after passing through the target is:

$v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )$

$v_{a,f} = 30.1\,\frac{m}{s}$

The speed of the arrow after passing through the target is 30.1 meters per second.

4 0

3 years ago