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3 years ago

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Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

f the trip would take him 16 hours longer than it would if he increased his speed by 50%. What is his current speed?

1 answer:

Verdich [7]3 years ago

6 0

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

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<h3>When acceleration is not zero, can speed remain constant?</h3>

The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.

Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.

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1 year ago

What is the resistance (R) when voltage is 179V and current is 5 Amps?

Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

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2 years ago

Someone please quickly help me with this problem?

Crazy boy [7]

The answer is 60 km. I hope it helps i dont know if this is right or wrong.

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3 years ago

Read 2 more answers

A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

MissTica

Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

5 0

3 years ago

A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

$B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

$B=6.978*10^{-5}T$

Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

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

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

8 0

2 years ago