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3 years ago

A couple walking in the park strolls 1.5 km in 60 min. What is the couple’s average speed in m/h?

Physics

2 answers:

cricket20 [7]3 years ago

6 0

Answer:

0.932 m/h

Explanation:

Parameters given:

Distance walked by couple, D = 1.5 km

Time spent in walking, t = 60 min

To solve this we simply apply the formula for average peed:

Average Speed = total distance / total time taken

The question asks specifically to put the answer in miles per hour (m/h), hence, we need to convert the distance to miles and the time to hours:

Distance in miles:

1 km = 0.622 miles

1.5 km = 1.5 * 0.622 = 0.932 miles

Time in hours = 60 / 60 = 1 hour

Hence, the speed is:

Speed = 0.932 / 1 = 0.932 m/h

The average speed of the couple is 0.932 m/h

ch4aika [34]3 years ago

6 0

Answer:

1,500m/h

Explanation:

the information that we have is:

distance traveled: $d=1.5km$

to convert this quantity to meters we remember that 1km = 1,000m

so the distance in meters is:

$d=1.5km(1000m/1km)\\d=1.5(1000)\\d=1,500m$

and we also know the time

time: $t=60min$ , and since 60min = 1hour, the time in hours is:

$t=1h$

now we use the formula for the average speed:

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

and we sustitute the known values (<u>since we want the answer to be in m/h we substitute the distance in meters (m) and the time in hours(h)</u>):

$v=\frac{1,500m}{1h}\\ v=1,500m/h$

the speed of the couple in m/h is: 1,500m/h

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3 years ago

Read 2 more answers

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$