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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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Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

      u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
The plunger of a syringe is pulled up halfway, as shown in the picture. With the plunger in this position, how does the volume o
noname [10]

Explanation. A gas will expand to fill its container. Pulling the plunger of the syringe creates a low pressure inside the syringe (a vacuum)

5 0
3 years ago
Read 2 more answers
How do our daily activities expose us to radiation?
dusya [7]

Explanation:

when we go outside, heat up food in the microwave, etc. it exposes us to radiation. along with things like radiators. the sun has lots of radiation and heat, so going outside exposes us. the microwave has radiation that heats up our food. the radiation from a microwave is not healthy for us, and is slowly poisoning us each day we use it. but as long as we do those things, we are exposed.

hope this helps

8 0
3 years ago
A bicycle pedal is pushed straight downwards by a foot with a 33 Newton force. The shaft of the pedal is 20 cm long. If the shaf
katen-ka-za [31]

Answer:

The magnitud of the torque doing by the foot at the point where the shaft is attached is 0.063 Nm

Explanation:

The torque is obtained by multiplying the longitude from the shaft attached point to force acting point by the force perpendicular component:

T = d * Fp

Notice that the perpindicular component is the total force magnitud times the sino of the angle respect the horizontal:

Fp = F*sin(a)

Replacing the values for the force and the angle:

Fp = 33N*sin(π/5)  = 33N * 0.011 = 0.363 N

Taking the distance in meters:

T = 0.2m * 0.363 N = 0.063 Nm

7 0
2 years ago
Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not cha
nordsb [41]

Answer : The temperature of water is, 57.7^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of thermometer = 804J/kg.^oC=0.804J/g^oC

c_2 = specific heat of water = 4.18J/g.^oC

m_1 = mass of thermometer = 33.0 g

m_2 = mass of water = 149 g

T_f = final temperature = 56.0^oC

T_1 = initial temperature of thermometer = 15.4^oC

T_2 = initial temperature of water = ?

Now put all the given values in the above formula, we get:

33.0g\times 0.804J/g^oC\times (56.0-15.4)^oC=-149g\times 4.18J/g.^oC\times (56.0-T_2)^oC

T_2=57.7^oC

Therefore, the temperature of water is, 57.7^oC

5 0
2 years ago
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