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Luden [163]
1 year ago
12

You drive 5 miles at 30 mi/hr and then another 3 miles at 50 mi/hr. What is your average speed for the whole8-mile trip?

Physics
1 answer:
s2008m [1.1K]1 year ago
6 0

Given data

The initial distance covered is d1 = 5 miles

The initial speed is s1 = 30 mi/hr

The final distance is d2 = 3 miles

The final speed is s2 = 50 mi/hr

The expression for the initial time taken to travel is given as:

t_1=\frac{d_1}{s_1}

The expression for the final time taken is given as:

t_2=\frac{d_2}{s_2}

The expression for the average speed for the whole trip is given as:

\begin{gathered} \text{Average sp}eed\text{ =}\frac{total\text{ distance}}{total\text{ time}} \\ S_{Avg}=\frac{d_1+d_2}{t_1+t_2} \\ S_{Avg}=\frac{d_1+d_2}{\frac{d_1}{s_1}+\frac{d_2}{s_2}_{}} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} S_{Avg}=\frac{5\text{ mi+3 mi}}{\frac{5\text{ mi}}{30\text{ mi/hr}}+\frac{3\text{ mi}}{50\text{ mi/hr}}} \\ S_{Avg}=35.3\text{ mi/hr} \end{gathered}

Thus, the average speed for the whole trip is 35.3 mi/hr.

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For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?
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4 0
3 years ago
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
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