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Andrews [41]
4 years ago
5

Complete this sentence. The solubility of a sample will ____________ when the size of the sample increases.

Physics
1 answer:
Murljashka [212]4 years ago
6 0
The correct answer is B.
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A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   We have equation of motion, s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate time when displacement = 1.5 meter.

 1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 m/s^2

Substituting

     -1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0

In case of horizontal motion

    Horizontal speed of car = V cos 7 = 0.993V

    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation 4.9t^2-0.122Vt-1.5=0

    We will get 4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

7 0
4 years ago
As an object increases or decreases, what happens to the total energy?
kompoz [17]

Answer:As an object speeds up, the total energy increases / decreases / remains the same. As an object slows down, the total energy increases / decreases / remains the same.

Explanation:

7 0
3 years ago
Please help will give branliest
USPshnik [31]

Nabr is the answer bcoz in a chemical reaction reactants react to form new substances and according to your options NaBr should be the answer

6 0
3 years ago
Which person below is experiencing a delusion?
Yakvenalex [24]

Answer: D Demetrius who sees distortions and colors when he looks at a face.

Explanation:

3 0
3 years ago
Read 2 more answers
A 110 kg football player running with a velocity of 5.0 m/s hits another stationary football
KengaRu [80]

Answer:

The final velocity of the second player is 6.1 m/s.

Explanation:

The final velocity of the second player can be calculated by conservation of linear momentum (p):

p_{i} = p_{f}  

m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}}  (1)

Where:

m_{a}: is the mass of the first football player = 110 kg

m_{b}: is the mass of the second football player = 90 kg

v_{a_{i}}: is the initial velocity of the first football player = 5.0 m/s

v_{b_{i}}: is the initial velocity of the second football player = 0 (he is at rest)

v_{a_{f}}: is the final velocity of the first football player = 0 (he stops after the impact)

v_{b_{f}}: is the final velocity of the second football player =?

By solving equation (1) for v_{b_{f}} we have:

110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}}

v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s

Therefore, the final velocity of the second player is 6.1 m/s.

I hope it helps you!

8 0
3 years ago
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