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worty [1.4K]
3 years ago
5

Rewrite this formula to solve for GGi: DX=fL/GGi

Physics
1 answer:
sashaice [31]3 years ago
3 0
Comment if you need clarification!

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A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what
Fudgin [204]

Answer:

v = 40 [m/s].

Explanation:

Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

P=m*v\\

where:

m = mass [kg]

v = velocity [m/s]

P =20*10\\P =200 [kg*m/s]

Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.

v = P/m\\v = 200/5\\v = 40 [m/s]

3 0
2 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
3 years ago
Are you a negro can i call you one
Roman55 [17]

Answer:

no it is mean call people their names

Explanation:

because if my name was Shar i would want people to call me Shar not negro

3 0
2 years ago
What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?
Stolb23 [73]
Speed = distance/time
speed= 122÷27=4.52m/s (3sf)
7 0
3 years ago
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30
dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

<em><u>Support Cy:</u></em>

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

<em><u>Support Ay:</u></em>

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

4 0
3 years ago
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