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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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Ganezh [65]

Answer:

Your answer would be A.

Explanation:

A. the rope will not move

3 0
3 years ago
Consider a simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d.
Aleksandr-060686 [28]

Answer:

c. Smaller than.

Explanation:

The energy stored in a capacitor is given as

E = 1/2CV²...................... Equation 1

Where E = Energy stored in a capacitor, C = capacitance of the capacitor, V = Voltage.

Also,

C = εA/d ......................Equation 2

Where ε = permitivity of the material, A = cross-sectional area of the plate, d = distance of separation of the plate.

substitute equation 2 into equation 1

E = 1/2εAV²/d ................. Equation 3

From equation 3 above,

The energy stored in a capacitor is inversely proportional to the distance of separation between the plates.

Hence when the plate is pulled apart by a distance D (D>d) The energy stored in the capacitor will be smaller.

The right option is c. Smaller than.

6 0
3 years ago
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric
gayaneshka [121]

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

KE = \fract{1}{2}mv^2

PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

m = 910 Kg

r_1 = 1200 + 6371 km = 7571km

r_2 = 6371 km,

Replacing we have,

\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})

v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})

v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})

v = 4456 m/s

Therefore the speed of the object when striking the surface of earth is 4456 m/s

3 0
4 years ago
Give your answer to 2 dp
AVprozaik [17]

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5

6 0
3 years ago
a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig
suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}

Then, the final velocity will be

v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s

Thus,  the final velocity of the truck is found as 146.969 m/s.

8 0
3 years ago
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