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4 years ago

5

The fiugure below shows a birds eye view of ta car going around a highway curve

1 answer:

luda_lava [24]4 years ago

5 0

Okay whats the question though?

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the average speed of a runner in a 483. meter race is 3.0 meters per second. How long me runner to complete the race? Dont inclu

lara [203]

Answer:

161

Explanation:

$v=\frac{d}{t}$ slove for t

$t=\frac{d}{v}$

Insert values of d and v

$t=\frac{483}{3} \\$

t=161

3 0

3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

Two waves collide and the temporary combined waves that results is smaller that the original wave.

zimovet [89]

A.

Destructive interference is when two waves cancel each other out or when the crest of one wave passes through the trough of another wave.

5 0

3 years ago

Which rule(s) is used for combining velocities?

DIA [1.3K]

Answer: B. Vector rules

Explanation:

4 0

3 years ago

A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross

Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

$F_D=\frac{1}{2}\rhoC_dAV^2$

Our values are:

$V=30.1m/s$

$C_d=0.45$

$A=3.3m^2$

$\rho=1.2kg/m^3$

Replacing,

$F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2$

$F_D=807.25N$

We need calculate now the energy lost through a time T, then,

$W = F_D d$

But we know that d is equal to

$d=vt$

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

$W=(807.25N)(30.1m/s)(3600s/1hr)$

$W=87.47*10^6J$ (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0

3 years ago