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inna [77]
4 years ago
6

You are watching a magic show. For one trick the magician rolls a ball down a hill. Suddenly the ball stops moving down the hill

.It is as if the ball is defying gravity! Come up with the It explanation for how the magician might have accomplished his trick. Hint: Think of all the forces that might on the ball.
Physics
2 answers:
poizon [28]4 years ago
8 0
The forces (what causes the ball to accelerate) are gravity, friction, and the normal force. In this case, gravity is a downward force caused by the gigantic mass of the Earth and the mass of the ball. Keep in mind that a force is acceleration. Acceleration is a change in velocity. The ball speeds up. Than it stops speeding up at a certain point where the frictional force (along with air friction) equals the parallel component of gravity. 




Newton's Second Law States- The greater mass of an object, the more force it will take to accelerate the object.


olchik [2.2K]4 years ago
6 0
The forces are gravity, friction, and normal force.
Hope this helps!!
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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Inessa05 [86]

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power  ,

initial speed of the car, v_0=0

final speed of the car, v=32 mph

At full power,

initial speed of the car, v_0=0

final speed of the car, v=64 mph

a)

At constant power, KE = \frac{1}{2} mv^2

At full power, KE = \frac{1}{2} m(2v)^2

So KE_f = 4KE_i

So, time to reach 64 mph speed is 4 times more than the initial time

t = 4*1.10 =4.40 s

b)

v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27 miles/hours^2

For final 64 mph speed,

v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4} hours = 6.111*10^{-4}*3600=2.20 s

7 0
3 years ago
We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat
Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)

Here, mw is the mass of water

Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal

Now, you need to calculate the heat released by the steam:

Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal

The work done by this engine is the difference between both heats:

W=Q_{1}-Q_{2}=62760-61960=800cal

8 0
3 years ago
How can briquette support to save fossil fuel?​
Marina86 [1]

Answer:

use a reusable source of energy such as hydro power

Explanation:

use renewable energy it saves the environment and does not create global warming

6 0
2 years ago
A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 41,200 ly (light-years) from the galactic ce
Ronch [10]

The total mass of the galaxy is 443.4 Solar mass

Orbital velocity (v) = \sqrt{\frac{MG}{R} }

where M= weight of galaxy

G= gravitational constatnt = 6.674*10^-^1^1 (given)

R = distance from centre = 41200 Light years (given)= 4.12*9.5*10^1^6  km (1 ly= 9.5*10^3 billion km)

v= orbital velocity = 275  km/s (given)

∴ According to the formula

(2.75*10^2)^2 = \frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}

⇒ 7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1 (cross multiplying and expanding)

⇒ 29.59*10^2^1=M*6.674*10^-^1^1

⇒ \frac{29.59*10^2^1*10^1^1}{6.674}=M

⇒ 4.434*10^3^2=M

1 solar mass = 1.989*10^3^0 kg

⇒ Mass in solar mass =443.4 Solar mass

⇒ M = 443.4 Solar mass

Learn more about Orbital velocity here :

brainly.com/question/22247460

#SPJ10

6 0
2 years ago
What is the minimum speed the rock must have at the top of the circle if it is to always stay in contact with the bottom of the
Naddika [18.5K]
 <span>First lets determine the equation. Well at the top of the circle both the normal force and the weight are in the same direction. So we have Fnet=N+mg. Since this is a circular path the Fnet is also = to (mv^2)/r. 

We convert the situation where the rock is no longer in contact with the bottom to terms relevant to the equation. So, what is a requirement for normal force? The object must be in contact with the surface, meaning it can't be in free fall. Realizing this means that the instant when the object does not touch the bucket is where the normal force = 0. 

Now we have N+mg=(mv^2)/r where N=0 is the case we are interested in. This leaves 0+mg=(mv^2)/r 
Solve for v: 
v=(gr)^(1/2) or v=3.28m/s 
</span>
7 0
3 years ago
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