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3 years ago

How long does it take anya to cover the distance of 5.00 miles ?

1 answer:

5 0

Here is the full information about the question. Ilya and Anya each can run at a speed of 8.50mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
the calculation would be:

t = d / s 
t = 2.5 (half of the total distance) / 8.5 (speed of running) 
This is .294 hours which is about 1058s... 

for the walking part... 

t = d / s 
t = 2.5 / 3.5 
t = 5/7hours = 2571 s.

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Blababa [14]

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

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now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

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as we know that pressure changes with depth as per following equation

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here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

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$2(1.01 \times 10^5) = 1025 (9.81)(h)$

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5 0

3 years ago

A car travels in a straight line from a position 50 m to your right to a position 210 m to your right in 5 sec. a. What is the a

Verizon [17]

Answer:

a) The average velocity is 32 m/s

b) See the attached figure. Slope of the line = 32.

Graphic: a line that passes through the points (0; 50) and (5; 210)

Explanation:

a) The average velocity is calculated as the displacement over time:

v = ΔX / Δt

where

ΔX : displacement ( final position - initial position)

Δt : time (final time - initial time)

Considering the origin of the reference system as the position where the observer is:

ΔX = 210 m - 50 m = 160 m

Δt = 5 s

v = 160 m / 5 s = 32 m/s

b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:

slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.

4 0

3 years ago

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

Gnom [1K]

Answer:

(a) Acceleration of positron is 6.03 x 10¹³ m/s²

(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s

Explanation:

Given :

Constant electric field, E = 343 N/C

Mass of positron, m = 9.1 x 10⁻³¹ kg

Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C

(a) Coulomb force on the positron is determine by the relation :

F = q x E ....(1)

But, force is also equals to product of mass and acceleration. So,

F = ma .....(2)

Here a is acceleration.

From equation (1) and (2).

m x a = q x E

$a=\frac{qE}{m}$

Substitute the values of q, E and m in the above equation.

$a=\frac{1.6\times10^{-19}\times 343}{9.1\times10^{-31} }$

a = 6.03 x 10¹³ m/s²

(b) Initially, the positron is at rest, so its initial speed is zero.

The equation of motion for positron is :

v = u + at

Here v is final speed, u is initial speed and t is time.

Since, u is zero, so the equation becomes :

v = at

Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.

v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s

v = 5.24 x 10⁵ m/s

6 0

3 years ago

(a) Calculate the linear acceleration of a car, the 0.220-m radius tires of which have an angular acceleration of 11.0 rad/s2. A

Alisiya [41]

Answer:

The value is $a_t = 2.42 \ m/s^2$

Explanation:

From the question we are told that

The radius of the tires is $r = 0.22 \ m$

The angular acceleration is $\alpha = 11.0 \ rad/s^2$

Generally the linear acceleration is mathematically represented as

$a_t = r * \alpha$

=> $a_t = 0.22 * 11$

=> $a_t = 2.42 \ m/s^2$

8 0

3 years ago

A car travels in a flat circle of radius R. At a certain instant the velocity of the car is 24 m/s west and the total accelerati

lisov135 [29]

Answer with Explanation:

We are given that

Velocity,v=24 m/s west

Acceleration,a= $2.5m/s^2$

$\theta=53^{\circ}$ (N of W)

Horizontal component of acceleration=Tangential acceleration

Tangential acceleration, $a_x=acos\theta=2.5cos53=1.505 m/s^2$

Radial acceleration=Vertical acceleration= $a_y=asin\theta$

$a_y=2.5sin53=1.998 m/s^2$

Radial acceleration, $a_y=\frac{v^2}{R}$

$1.998=\frac{(24)^2}{R}$

$R=\frac{(24)^2}{1.998}$

$R=288.29 m$

Time, $t=\frac{2\pi R}{velocity}$

$t=\frac{2\pi(288.29)}{24}$

$t=75.47 s$

8 0

3 years ago