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2 years ago

How rolling friction is less than sliding friction?

1 answer:

6 0

Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

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Can someone help me ?

hammer [34]

Answer:

1) Time interval Blue Car Red Car

0 - 2 s Constant Velocity Increasing Velocity

2 - 3 s Constant Velocity Constant Velocity

3 - 5 s Constant Velocity Increasing Velocity

5 - 6 s Constant Velocity Decreasing Velocity

2) For Red and Blue car y₂ = 120 v = $\frac{y_{2}-y_{1}}{t_{2}-t_{1}}$ = $\frac{120-0}{6-0}$ = 20 m/s

We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.

3) At t = 2s, the cars are the same position, and are moving at the same rate

Position - same

Velocity - same

The position-time graph shares the same spot for two cars.

4 0

3 years ago

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

$v^2=u^2+2as$

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

$0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s$

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

$v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s$

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

$v=u+at$

where symbols have the usual meaning

Applying the given values we get

$t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds$

4 0

2 years ago

The current in an electric hair dryer is 10

Vikki [24]

First let's convert the time in seconds:
$\Delta t= 5 min= 5 min \cdot (60 s/min)= 300 s$

The current is defined as the quantity of charge flowing through a certain section of a circuit per unit of time:
$I= \frac{Q}{\Delta t}$
Using I=10 A, and $\Delta t=300 s$ , we can find the amount of charge flown through the hair dryer in this time:
$Q=I \Delta t=(10 A)(300 s)=3000 C$

The charge of a single electron is $q=1.6 \cdot 10^{-19} C$ , so the number of electrons flown through the hair dryer is the total charge divided by the charge of a single electron:
$N= \frac{Q}{q}= \frac{3000 C}{1.6 \cdot 10^{-19} C} =1.88 \cdot 10^{22}$

8 0

3 years ago

julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a

aliya0001 [1]

Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2
Time to height
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)

8 0

3 years ago

Static cling is an example of

cestrela7 [59]

I would say c hope it helps:)

3 0

3 years ago