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pantera1 [17]
3 years ago
12

Suppose the designers of the water slide want to adjust the height h above the water so that riders land twice as far away from

the bottom of the slide. What would be the necessary height above the water?
A. 1.2 m
B. 1.8 m
C. 2.4 m
D. 3.0 m
Physics
1 answer:
Nitella [24]3 years ago
4 0

Answer:

option C

Explanation:

given,

Range is doubled

for range to be doubled height should become 4 times.

formula to calculate the horizontal range

          R = v T_{flight}

using equation of motion

s = u t + \dfrac{1}{2}at^2

t=\sqrt{\dfrac{2s}{g}}

now,

      R = v\sqrt{\dfrac{2h}{g}}

where, h is the height of the

Assuming the height be equal to 0.6 m

to calculate the new height

         h_{new}= 4 h

         h_{new}= 4\times 0.6

        h_{new}= 2.4\ m

hence, the correct answer is option C

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A piece of putty and a tennis ball with the same mass are thrown against a wall with the same velocity. Which object experiences
cluponka [151]

Answer:

Explanation:

Firstly, we have to define momentum.

Momentum is define as the product of mass and velocity.

That is P = mass×velocity

Also considering the third law of motion which states that: For every action, there is equal and opposite reaction.

Moreso, considering the 2nd law of motion which states that the rate of change in the momentum of a body is equal to the applied force and takes place in the direction of the applied force.

Now, applying P = mass×velocity

They both have same mass and velocity definitely, they will both experience same momentum.

Also from the question, the both share same velocity hence, the will both hit the wall with same velocity meaning the will both feel the same impact from the wall as well. Hence the third law of motion proves this right.

5 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
4 years ago
What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed
Pani-rosa [81]

     charge must be equal to 5.74 ×10⁻⁵

 In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

     →    Fnet =0

     →    mg =  qE

 substituting the values we get :

         0.00345 × 9.81 =  q × 590

   →       q = 5.74 ×10⁻⁵

    Hence the charge must be equal to   5.74 ×10⁻⁵.

   Learn more about charges here:

          brainly.com/question/26092261

                    # SPJ4

8 0
2 years ago
11) A sled is initially given a shove up a frictionless 35º incline. It reaches a maximum height of 2.5
Andrej [43]

Answer:

7 m/s

Explanation:

To solve this problem you must use the conservation of energy.

K1 +U1=K2+U2

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.

The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:

KE1= \frac{1}{2}mv^{2}}

PE2=mgh

K1=U2

\frac{1}{2} mv^{2} =mgh

Solve for v

v=\sqrt{2gh}

Input known values and you get 7 m/s.

5 0
3 years ago
What is the half-life for the radioactive decay of particle that
pogonyaev

Answer:

ytyug

Explanation:

4 0
3 years ago
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