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3 years ago

9

One way to ensure that the race ends up in a tie is to have the person who lost the original race run a shorter distance before

turning around. Instead of 80.0 m, what should that distance be

1 answer:

Vikki [24]3 years ago

6 0

Answer:

The appropriate answer is "72.8 m".

Explanation:

Let the distance be "x".

As per the question, we get

⇒ $T_{Mia}=T_{Brandi}$

or,

⇒ $\frac{x}{13} +\frac{x}{7}=\frac{\80}{10} +\frac{80}{10}$

By taking L.C.M, we get

⇒ $\frac{20x}{91}=16$

By applying cross-multiplication, we get

⇒ $20x=1440$

⇒ $x=\frac{1456}{20}$

⇒ $=72.8\ m$

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What is the internal resistor of the cell in closed circuit?

Drupady [299]

Generally, the internal resistance of the new battery is small, about 0.2 euros, while the old battery is large, close to 1 euro ，

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4 years ago

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

strojnjashka [21]

Answer:

$D=99.4665307m \approx 99.5m$

Explanation:

From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

6 0

3 years ago

A circular conducting loop of radius 23.0 cm is located in a region of homogeneous magnetic field of magnitude 0.500 T pointing

Alchen [17]

Answer:

The magnitude of the induced emf is

Ф = 5.419 x 10⁻³ V

Explanation:

Given:

r = 23.0 cm = 0.23 m

β₁ = 0.50 T

β₂ = 2.50 * 0.50 T = 1.25 T

Calculate the magnitude of the induced emf the magnetic field is increasing

Ф = A * Δβ / Δt

A = π * r² = π * 0.23² m = 0.166 m²

Ф = 0.166m² * (1.25 - 0.5) T / 23 s

Ф = 5.419 x 10⁻³ V

4 0

4 years ago

Im confused can someone explain what I have to do

Studentka2010 [4]

For numbers 1 and 3 it is giving you the answers i believe

number 1 would be 20 joules of potential energy on the far left and far right point

number 3 would the (3rd) point as it says it has 20 joules of kinetic energy

sorry if this doesn’t help

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3 years ago

Two unknown resistors are connected together. When they are connected in series their equivalent resistance is 15 Ω. When they a

DochEvi [55]

Explanation:

Let x and y are two unknown resistors. When they are connected in series their equivalent resistance is 15 Ω. When they are connected in parallel, their equivalent resistance is 3.3 Ω.

For series combination,

$x+y=15$ ......(1)

For parallel combination,

$\dfrac{1}{x}+\dfrac{1}{y}=3.3$ ....(2)

We need to find the resistances of these resistors. Solving equation (1) and (2) we get :

x = 0.29 and y = 14.7

Hence, the resistances of these resistors are 0.29 ohms and 14.7 ohms.

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3 years ago