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Stells [14]
3 years ago
12

Galileo started modern science because he started using ? Rather than speculation

Physics
1 answer:
Eva8 [605]3 years ago
4 0
By using the telescope
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A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10
Alchen [17]

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

T = 39 days \times 24 hr \times 3600 s

T = 3369600 s

now the radius of the orbit is given as

r = 4.8 \times 10^8 m

so the total path length is given as

L = 2 \pi r

L = 2\pi (4.8 \times 10^8)

L = 3.015 \times 10^9

now the speed will be given as

v = \frac{L}{T}

v = \frac{3.015 \times 10^9}{3369600}

v = 895 m/s

7 0
3 years ago
This is a Science question pls help give 15 coins for answer
Vlad1618 [11]

Answer:

<u>Conservation</u>: using less water

<u>Xeriscaping</u>: replanting your yard with plants that do not require great amounts of water

<u>Desalination</u>: process of removing salt from water so that it can be used for consumption

<u>Water Budget</u>: finite amount of usable water available

<u>Potable</u>: water that is safe to use a drink

7 0
3 years ago
Read 2 more answers
If the frequency of the incoming light is increased, will the energy of the ejected electrons increase, decrease, or stay the sa
Arlecino [84]

Answer:

increase

Explanation:

According to Einstein's photoelectric equation; the energy of a photon striking a metal surface is related to the kinetic energy of the ejected photoelectron by the formula;

KE= hf - hfo

Where h is the planks constant, f and fo refer to the frequency of incident photon and the threshold frequency respectively.

Hence, we can clearly see from the foregoing that the kinetic energy of the ejected photoelectron is proportional to the frequency of the incident photon.

Hence, if the frequency of the incident photon is increased, the kinetic energy of the ejected photoelectron increases also.

3 0
3 years ago
A 4 cm diameter ball is located 40 cm from a point source and 80 cm from a wall. What is the size of the shadow on the wall?
Slav-nsk [51]
<span>11.823 cm There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall. The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm. Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So 82 + 1/21 * 2 = 82 + 2/21 = 82.0952381 Now we have the following dimensions with a circle replacing the ball in the original problem. Distance from wall to effective circle = 82.0952381 cm Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm Effective diameter of circle = 3.995462279 cm And because the geometry makes similar triangles, the following ratio applies. 3.995462279/41.9047619 = X/124 Now solve for X 3.995462279/41.9047619 = X/124 124*3.995462279/41.9047619 = X 495.4373226/41.9047619 = X 11.82293611 = X The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
7 0
3 years ago
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