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2 years ago

9

Ramu,the gardener,is trying to pull out weeds. however,he has to apply great force.why do you think he has to apply to much forc

e?

1 answer:

jasenka [17]2 years ago

4 0

Answer:

Roots are basically hooked into the ground?

Explanation:

Maybe you could comment here the choices if it's multiple choice.

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A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2$

b)

Also , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m$

Hence , this is the required solution .

6 0

3 years ago

1. A limiting factor for using nuclear energy is the

Goryan [66]

Answer:

waste

Explanation:

took da test. nhgvgg

8 0

3 years ago

Read 2 more answers

A boat cruises 60 meters west in 10 seconds. What is its instantaneous velocity?

kari74 [83]

A - 6 meters a second ( 6m/s)

3 0

3 years ago

Read 2 more answers

In the circuit seen here, the resistor has a resistance of 3 ohms. If no change in the battery size occurs, what will happen to

Bumek [7]

The current will decrease as the resistance has now increased, meaning less current will be 'let through' the resistor. (assuming it's in series, there's no image)

6 0

3 years ago

Read 2 more answers

The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

$\frac{P_{1} }{\\P_2}$ = $\frac{density * g * h_1}{density * g * h_2}$

So $\frac{39}{P_2}$ = $\frac{3}{9}$

Or P₂= 39 * 3 = 117 kPa

5 0

3 years ago