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Andreyy89
4 years ago
14

Ladybug walks 10 cm forward and 5 cm backwards in 20 seconds what is the average speed of ladybug what is the average velocity

Physics
2 answers:
Tamiku [17]4 years ago
4 0

Hey!

---------------------------------------------------

Formula: Distance / Time = Speed

---------------------------------------------------

Solution:

Speed = 15/20 = 0.75cm/sec

Velocity = 5/20 = 0.25cm/sec

10 + 5 = 15 (distance traveled)

---------------------------------------------------

Answer:

Speed = 0.75cm/sec

Velocity = 0.25cm/sec going forward.

---------------------------------------------------

Hope This Helped! Good Luck!

lbvjy [14]4 years ago
3 0

Answer:

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

Explanation:

Speed is the ratio of distance and time and velocity is the ratio of displacement and time.

Distance traveled = 10 + 5 = 15 cm = 0.15 m

Displacement = 10 - 5 = 5 cm = 0.05 m

Time = 20 seconds

\texttt{Average speed = }\frac{\texttt{Distance}}{\texttt{Time}}\\\\\texttt{Average speed = }\frac{0.15}{20}=7.5\times 10^{-3}m/s\\\\\texttt{Average velocity = }\frac{\texttt{Displacement}}{\texttt{Time}}\\\\\texttt{Average velocity = }\frac{0.05}{20}=2.5\times 10^{-3}m/s

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

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A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +46 N·m is ap
Rom4ik [11]

Answer:

Explanation:

Let Torque due to friction be

F  

Net torque

= 46 - F

Angular impulse = change in angular momentum

=(  46 - F ) x 17  = I X 580

When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so

Angular impulse = change in angular momentum

F  x 120 = I X 580

(  46 - F ) x 17 = F  x 120

137 F = 46 x 17

F = 5.7 Nm

b )

Putting this value in first equation

5.7 x 120 = I x 580

I = 1.18 kg m²

8 0
3 years ago
For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0
3 years ago
The world's fastest production sportscar has a top speed of 415 kmh-1(a)Convert this speed to ms-1.[ 1](b)The distance from Lond
Alex73 [517]

Answer:

(a). The speed is 115.28 m/s.

(b). The time is 94 min.

(c). The International Space Station travels 43202.400 km.

(d). Speed is scalar quantity.

Velocity is vector quantity.

Explanation:

Given that,

Top speed = 415 km/h

1 miles = 1609 m

(a). We need to calculate the speed in m/s

Using conversion of km/h to m/s

v=415\ km/h

v=415\times\dfrac{5}{18}

v=115.28\ m/s

(b). We need to calculate the distance from London to Edinburgh in km

Using conversion for distance

d=403\ miles

Distance in meter,

d=403\times1609

d=648427\ m

d=648.427\ km

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{648.427}{415}

t=1.56\ h

t=1\ hours 34\ min

t= 94\ min

(c). Speed v' =7.66 km/s

v'=7660 m/s

In the time it takes the car to travel from London to Edinburgh,

We need to calculate the distance

Using formula of distance

d'=v'\times t

Put the value into the formula

d'=7660\times94\times60

d'=43202400\ m

d'=43202.400\ km

(d). Speed :

Speed is equal to the distance divided by time.

It is scalar quantity.

Velocity :

Velocity is equal to the displacement divided by time.

It is vector quantity.

Hence, (a). The speed is 115.28 m/s.

(b). The time is 94 min.

(c). The International Space Station travels 43202.400 km.

(d). Speed is scalar quantity.

Velocity is vector quantity.

3 0
3 years ago
A mass of 1 kg is moving in a circle of radius 3.3 meters. What is the liner velocity v m/s that would give a Centripetal Force
Ghella [55]

Answer:

The linear velocity of the object is 8.71 m/s.

Explanation:

Given;

mass of the object, m = 1 kg

radius of the circle, r = 3.3 meters

centripetal force, F = 23 N

Centripetal force is given by;

F_c = \frac{mv^2}{r}\\\\

where;

v is the linear velocity of the object

F_c = \frac{mv^2}{r}\\\\mv^2 = F_cr\\\\v^2 = \frac{F_cr}{m}\\\\v= \sqrt{\frac{F_cr}{m}} \\\\v= \sqrt{\frac{23*3.3}{1}}\\\\v = 8.71 \ m/s

Therefore, the linear velocity of the object is 8.71 m/s.

4 0
3 years ago
Which has more gravitaatoinal potental energy: a bird on the ground or the same bird in the tree? why?
julsineya [31]
The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.

Considering also the formula for Gravitational Potential Energy GPE = mgh

For the bird on the ground, h =0,  therefore GPE = m*9.8*0 = 0

For that on the tree = mgh = m*9.8*h

Of course the one on the tree has a value greater than zero.  
7 0
3 years ago
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